# Number Theory for Mathematical Contests. David A. SANTOS

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Number Theory for Mathematical Contests David A. SANTOS August 13, 2005 REVISION

2 Contents Preface 1 Preliminaries Introduction Well-Ordering Practice Mathematical Induction Practice Fibonacci Numbers Practice Pigeonhole Principle Practice Divisibility Divisibility Practice Division Algorithm Practice Some Algebraic Identities Practice Congruences. Z n Congruences Practice Divisibility Tests Practice Complete Residues Practice Unique Factorisation GCD and LCM Practice Primes Practice Fundamental Theorem of Arithmetic Practice iii 5 Linear Diophantine Equations Euclidean Algorithm Practice Linear Congruences Practice A theorem of Frobenius Practice Chinese Remainder Theorem Practice Number-Theoretic Functions Greatest Integer Function Practice De Polignac s Formula Practice Complementary Sequences Practice Arithmetic Functions Practice Euler s Function. Reduced Residues Practice Multiplication in Z n Practice Möbius Function Practice More on Congruences Theorems of Fermat and Wilson Practice Euler s Theorem Practice Scales of Notation The Decimal Scale Practice Non-decimal Scales Practice A theorem of Kummer Miscellaneous Problems 91 Practice

3 Preface These notes started in the summer of 1993 when I was teaching Number Theory at the Center for Talented Youth Summer Program at the Johns Hopkins University. The pupils were between 13 and 16 years of age. The purpose of the course was to familiarise the pupils with contest-type problem solving. Thus the majority of the problems are taken from well-known competitions: AHSME AIME USAMO IMO ITT MMPC (UM) 2 STANFORD MANDELBROT American High School Mathematics Examination American Invitational Mathematics Examination United States Mathematical Olympiad International Mathematical Olympiad International Tournament of Towns Michigan Mathematics Prize Competition University of Michigan Mathematics Competition Stanford Mathematics Competition Mandelbrot Competition Firstly, I would like to thank the pioneers in that course: Samuel Chong, Nikhil Garg, Matthew Harris, Ryan Hoegg, Masha Sapper, Andrew Trister, Nathaniel Wise and Andrew Wong. I would also like to thank the victims of the summer 1994: Karen Acquista, Howard Bernstein, Geoffrey Cook, Hobart Lee, Nathan Lutchansky, David Ripley, Eduardo Rozo, and Victor Yang. I would like to thank Eric Friedman for helping me with the typing, and Carlos Murillo for proofreading the notes. Due to time constraints, these notes are rather sketchy. Most of the motivation was done in the classroom, in the notes I presented a rather terse account of the solutions. I hope some day to be able to give more coherence to these notes. No theme requires the knowledge of Calculus here, but some of the solutions given use it here and there. The reader not knowing Calculus can skip these problems. Since the material is geared to High School students (talented ones, though) I assume very little mathematical knowledge beyond Algebra and Trigonometry. Here and there some of the problems might use certain properties of the complex numbers. A note on the topic selection. I tried to cover most Number Theory that is useful in contests. I also wrote notes (which I have not transcribed) dealing with primitive roots, quadratic reciprocity, diophantine equations, and the geometry of numbers. I shall finish writing them when laziness leaves my weary soul. I would be very glad to hear any comments, and please forward me any corrections or remarks on the material herein. David A. SANTOS iii

4 Legal Notice This material may be distributed only subject to the terms and conditions set forth in the Open Publication License, version 1.0 or later (the latest version is presently available at THIS WORK IS LICENSED AND PROVIDED AS IS WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IM- PLIED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE OR A WARRANTY OF NON-INFRINGEMENT. THIS DOCUMENT MAY NOT BE SOLD FOR PROFIT OR INCORPORATED INTO COMMERCIAL DOCUMENTS WITHOUT EXPRESS PERMISSION FROM THE AUTHOR(S). THIS DOCUMENT MAY BE FREELY DISTRIBUTED PROVIDED THE NAME OF THE ORIGINAL AUTHOR(S) IS(ARE) KEPT AND ANY CHANGES TO IT NOTED. iv

5 Chapter1 Preliminaries 1.1 Introduction We can say that no history of mankind would ever be complete without a history of Mathematics. For ages numbers have fascinated Man, who has been drawn to them either for their utility at solving practical problems (like those of measuring, counting sheep, etc.) or as a fountain of solace. Number Theory is one of the oldest and most beautiful branches of Mathematics. It abounds in problems that yet simple to state, are very hard to solve. Some number-theoretic problems that are yet unsolved are: 1. (Goldbach s Conjecture) Is every even integer greater than 2 the sum of distinct primes? 2. (Twin Prime Problem) Are there infinitely many primes p such that p+2 is also a prime? 3. Are there infinitely many primes that are 1 more than the square of an integer? 4. Is there always a prime between two consecutive squares of integers? In this chapter we cover some preliminary tools we need before embarking into the core of Number Theory. 1.2 Well-Ordering The set N = {0,1,2,3,4,...} of natural numbers is endowed with two operations, addition and multiplication, that satisfy the following properties for natural numbers a,b, and c: 1. Closure: a+b and ab are also natural numbers. 2. Associative laws: (a+b)+c = a+(b+c) and a(bc) = (ab)c. 3. Distributive law: a(b+c) = ab+ac. 4. Additive Identity: 0+a = a+0 = a 5. Multiplicative Identity: 1a = a1 = a. One further property of the natural numbers is the following. 1 Axiom (Well-Ordering Axiom) Every non-empty subset S of the natural numbers has a least element. As an example of the use of the Well-Ordering Axiom, let us prove that there is no integer between 0 and 1. 2 Example Prove that there is no integer in the interval ]0;1[. 1

6 2 Chapter 1 Solution: Assume to the contrary that the set S of integers in ]0;1[ is non-empty. Being a set of positive integers, it must contain a least element, say m. Now, 0 < m 2 < m < 1, and so m 2 S. But this is saying that S has a positive integer m 2 which is smaller than its least positive integer m. This is a contradiction and so S =. We denote the set of all integers by Z, i.e., Z = {... 3, 2, 1,0,1,2,3,...}. A rational number is a number which can be expressed as the ratio a of two integers a,b, where b 0. We denote the set of b rational numbers by Q. An irrational number is a number which cannot be expressed as the ratio of two integers. Let us give an example of an irrational number. 3 Example Prove that 2 is irrational. Solution: The proof is by contradiction. Suppose that 2 were rational, i.e., that 2 = a for some integers a,b. This implies b that the set A = {n 2 : both n and n 2 positive integers} is nonempty since it contains a. By Well-Ordering A has a smallest element, say j = k 2. As 2 1 > 0, j( 2 1) = j 2 k 2 = ( j k) 2 is a positive integer. Since 2 < 2 2 implies 2 2 < 2 and also j 2 = 2k, we see that ( j k) 2 = k(2 2) < k( 2) = j. Thus ( j k) 2 is a positive integer in A which is smaller than j. This contradicts the choice of j as the smallest integer in A and hence, finishes the proof. 4 Example Let a,b,c be integers such that a 6 + 2b 6 = 4c 6. Show that a = b = c = 0. Solution: Clearly we can restrict ourselves to nonnegative numbers. Choose a triplet of nonnegative integers a, b, c satisfying this equation and with max(a,b,c) > 0 as small as possible. If a 6 + 2b 6 = 4c 6 then a must be even, a = 2a 1. This leads to 32a b6 = 2c 6. Hence b = 2b 1 and so 16a b6 1 = c6. This gives c = 2c 1, and so a b6 1 = 4c6 1. But clearly max(a 1,b 1,c 1 ) < max(a,b,c). This means that all of these must be zero. 5 Example (IMO 1988) If a,b are positive integers such that a2 + b 2 1+ab is an integer, then a2 + b 2 1+ab is a perfect square. Solution: Suppose that a2 + b 2 = k is a counterexample of an integer which is not a perfect square, with max(a,b) as small as 1+ab possible. We may assume without loss of generality that a < b for if a = b then 0 < k = 2a2 a < 2, which forces k = 1, a perfect square. Now, a 2 + b 2 k(ab+1) = 0 is a quadratic in b with sum of the roots ka and product of the roots a 2 k. Let b 1,b be its roots, so b 1 + b = ka and b 1 b = a 2 k. As a,k are positive integers, supposing b 1 < 0 is incompatible with a 2 + b 2 1 = k(ab 1 + 1). As k is not a perfect square, supposing b 1 = 0 is incompatible with a = k(0 a+1). Also b 1 = a2 k b < b2 k b < b.

7 Practice 3 Thus we have found another positive integer b 1 for which a2 + b ab 1 = k and which is smaller than the smallest max(a,b). This is a contradiction. It must be the case, then, that k is a perfect square. Practice 6 Problem Find all integer solutions of a 3 + 2b 3 = 4c 3. 7 Problem Prove that the equality x 2 +y 2 +z 2 = 2xyz can hold for whole numbers x,y,z only when x = y = z = Mathematical Induction The Principle of Mathematical Induction is based on the following fairly intuitive observation. Suppose that we are to perform a task that involves a certain number of steps. Suppose that these steps must be followed in strict numerical order. Finally, suppose that we know how to perform the n-th task provided we have accomplished the n 1-th task. Thus if we are ever able to start the job (that is, if we have a base case), then we should be able to finish it (because starting with the base case we go to the next case, and then to the case following that, etc.). Thus in the Principle of Mathematical Induction, we try to verify that some assertion P(n) concerning natural numbers is true for some base case k 0 (usually k 0 = 1, but one of the examples below shows that we may take, say k 0 = 33.) Then we try to settle whether information on P(n 1) leads to favourable information on P(n). We will now derive the Principle of Mathematical Induction from the Well-Ordering Axiom. 8 Theorem (Principle of Mathematical Induction) If a sets of non-negative integers contains the integer 0, and also contains the integer n+1 whenever it contains the integer n, then S = N. Proof: Assume this is not the case and so, by the Well-Ordering Principle there exists a least positive integer k not in S. Observe that k > 0, since 0 S and there is no positive integer smaller than 0. As k 1 < k, we see that k 1 S. But by assumption k 1+1 is also in S, since the successor of each element in the set is also in the set. Hence k = k 1+1 is also in the set, a contradiction. Thus S = N. The following versions of the Principle of Mathematical Induction should now be obvious. 9 Corollary If a set A of positive integers contains the integer m and also contains n+1 whenever it contains n, where n > m, then A contains all the positive integers greater than or equal to m. 10 Corollary (Principle of Strong Mathematical Induction) If a set A of positive integers contains the integer m and also contains n+1 whenever it contains m+1,m+2,...,n, where n > m, then A contains all the positive integers greater than or equal to m. We shall now give some examples of the use of induction. 11 Example Prove that the expression is a multiple of 169 for all natural numbers n. 3 3n+3 26n 27 Solution: For n = 1 we are asserting that = 676 = is divisible by 169, which is evident. Assume the assertion is true for n 1,n > 1, i.e., assume that 3 3n 26n 1 = 169N for some integer N. Then 3 3n+3 26n 27 = n 26n 27 = 27(3 3n 26n 1)+676n

8 4 Chapter 1 which reduces to N n, which is divisible by 169. The assertion is thus established by induction. 12 Example Prove that is an even integer and that (1+ 2) 2n +(1 2) 2n (1+ 2) 2n (1 2) 2n = b 2 for some positive integer b, for all integers n 1. Solution: We proceed by induction on n. Let P(n) be the proposition: (1+ 2) 2n +(1 2) 2n is even and (1+ 2) 2n (1 2) 2n = b 2 for some b N. If n = 1, then we see that (1+ 2) 2 +(1 2) 2 = 6, an even integer, and (1+ 2) 2 (1 2) 2 = 4 2. Therefore P(1) is true. Assume that P(n 1) is true for n > 1, i.e., assume that (1+ 2) 2(n 1) +(1 2) 2(n 1) = 2N for some integer N and that (1+ 2) 2(n 1) (1 2) 2(n 1) = a 2 for some positive integer a. Consider now the quantity (1+ 2) 2n +(1 2) 2n = (1+ 2) 2 (1+ 2) 2n 2 +(1 2) 2 (1 2) 2n 2. This simplifies to (3+2 2)(1+ 2) 2n 2 +(3 2 2)(1 2) 2n 2. Using P(n 1), the above simplifies to 12N + 2 2a 2 = 2(6N + 2a), an even integer and similarly (1+ 2) 2n (1 2) 2n = 3a 2+2 2(2N) = (3a+4N) 2, and so P(n) is true. The assertion is thus established by induction. 13 Example Prove that if k is odd, then 2 n+2 divides k 2n 1 for all natural numbers n. Solution: The statement is evident for n = 1, as k 2 1 = (k 1)(k+ 1) is divisible by 8 for any odd natural number k because both (k 1) and (k + 1) are divisible by 2 and one of them is divisible by 4. Assume that 2 n+2 k 2n 1, and let us prove that 2 n+3 k 2n+1 1. As k 2n+1 1 = (k 2n 1)(k 2n + 1), we see that 2 n+2 divides (k 2n 1), so the problem reduces to proving that 2 (k 2n + 1). This is obviously true since k 2n odd makes k 2n + 1 even.

9 Mathematical Induction 5 14 Example (USAMO 1978) An integer n will be called good if we can write n = a 1 + a 2 + +a k, where a 1,a 2,...,a k are positive integers (not necessarily distinct) satisfying 1 a a a k = 1. Given the information that the integers 33 through 73 are good, prove that every integer 33 is good. Solution: We first prove that if n is good, then 2n+8 and 2n+9 are good. For assume that n = a 1 + a 2 + +a k, and Then 2n+8 = 2a 1 + 2a a k and Also, 2n+9 = 2a 1 + 2a a k and Therefore, 1 = 1 a a a k. 1 2a a a k = = a 1 2a 2 2a k = = 1. if n is good both 2n+8 and 2n+9 are good. (1.1) We now establish the truth of the assertion of the problem by induction on n. Let P(n) be the proposition all the integers n,n+1,n+2,...,2n+7 are good. By the statement of the problem, we see that P(33) is true. But (1.1) implies the truth of P(n+1) whenever P(n) is true. The assertion is thus proved by induction. We now present a variant of the Principle of Mathematical Induction used by Cauchy to prove the Arithmetic-Mean- Geometric Mean Inequality. It consists in proving a statement first for powers of 2 and then interpolating between powers of Theorem (Arithmetic-Mean-Geometric-Mean Inequality) Let a 1,a 2,...,a n be nonnegative real numbers. Then n a1 a 2 a n a 1 + a 2 + +a n. n Proof: Since the square of any real number is nonnegative, we have ( x 1 x 2 ) 2 0. Upon expanding, x 1 + x 2 x 1 x 2, 2 (1.2) which is the Arithmetic-Mean-Geometric-Mean Inequality for n = 2. Assume that the Arithmetic-Mean-Geometric- Mean Inequality holds true for n = 2 k 1,k > 2, that is, assume that nonnegative real numbers w 1,w 2,...,w 2 k 1 satisfy w 1 + w 2 + +w 2 k 1 2 k 1 (w 1 w 2 w 2 k 1) 1/2k 1. (1.3) Using (1.2) with x 1 = y 1 + y 2 + +y 2 k 1 2 k 1 and x 2 = y 2 k y 2 k 2 k 1,

10 6 Chapter 1 we obtain that y 1 + y 2 + +y 2 k 1 2 k 1 + y 2 k y 2 k 2 k 1 2 Applying (1.3) to both factors on the right hand side of the above, we obtain ( y 1 + y 2 + +y 2 k 1 2 k 1 )( y 2 k y 2 k 2 k 1 ) 1/2. y 1 + y 2 + +y 2 k 2 k (y 1 y 2 y 2 k) 1/2k. (1.4) This means that the 2 k 1 -th step implies the 2 k -th step, and so we have proved the Arithmetic-Mean-Geometric- Mean Inequality for powers of 2. Now, assume that 2 k 1 < n < 2 k. Let and Let Using (1.4) we obtain which is to say that y 1 = a 1,y 2 = a 2,...,y n = a n, y n+1 = y n+2 = = y 2 k = a 1 + a 2 + +a n. n A = a 1 + +a n n and G = (a 1 a n ) 1/n. a 1 + a 2 + +a n +(2 k n) a 1 + +a n n 2 k a 1 a 2 a n ( a 1 + +a n n) 1/2 k ) (2k, n This translates into A G or which is what we wanted. na+(2 k n)a 2 k (G n A 2k n ) 1/2k. (a 1 a 2 a n ) 1/n a 1 + a 2 + +a n, n 16 Example Let s be a positive integer. Prove that every interval [s;2s] contains a power of 2. Solution: If s is a power of 2, then there is nothing to prove. If s is not a power of 2 then it must lie between two consecutive powers of 2, i.e., there is an integer r for which 2 r < s < 2 r+1. This yields 2 r+1 < 2s. Hence s < 2 r+1 < 2s, which gives the required result. 17 Example Let M be a nonempty set of positive integers such that 4x and [ x] both belong to M whenever x does. Prove that M is the set of all natural numbers. Solution: We will prove this by induction. First we will prove that 1 belongs to the set, secondly we will prove that every power of 2 is in the set and finally we will prove that non-powers of 2 are also in the set. Since M is a nonempty set of positive integers, it has a least element, say a. By assumption a also belongs to M, but a < a unless a = 1. This means that 1 belongs to M. Since 1 belongs to M so does 4, since 4 belongs to M so does 4 4 = 4 2, etc.. In this way we obtain that all numbers of the form 4 n = 2 2n,n = 1,2,... belong to M. Thus all the powers of 2 raised to an even power belong to M. Since the square roots belong as well to M we get that all the powers of 2 raised to an odd power also belong to M. In conclusion, all powers of 2 belong to M.

11 Practice 7 Assume now that n N fails to belong to M. Observe that n cannot be a power of 2. Since n M we deduce that no integer in A 1 = [n 2,(n + 1) 2 ) belongs to M, because every member of y A 1 satisfies [ y] = n. Similarly no member z A 2 = [n 4,(n+1) 4 ) belongs to M since this would entail that z would belong to A 1, a contradiction. By induction we can show that no member in the interval A r = [n 2r,(n+1) 2r ) belongs to M. We will now show that eventually these intervals are so large that they contain a power of 2, thereby obtaining a contradiction to the hypothesis that no element of the A r belonged to M. The function f : R + R x log 2 x is increasing and hence log 2 (n+1) log 2 n > 0. Since the function f : R R + x 2 x is decreasing, for a sufficiently large positive integer k we have 2 k < log 2 (n+1) log 2 n. This implies that (n+1) 2k > 2n 2k. Thus the interval [n 2k,2n 2k ] is totally contained in [n 2k,(n+1) 2k ). But every interval of the form [s,2s] where s is a positive integer contains a power of 2. We have thus obtained the desired contradiction. Practice 18 Problem Prove that 11 n n+1 is divisible by 133 for all natural numbers n. 19 Problem Prove that equals 1 x 1! + x(x 1) 2! x(x 1)(x 2) 3! + +( 1) n x(x 1)(x 2) (x n+1) n! ( 1) n (x 1)(x 2) (x n) n! for all non-negative integers n. 20 Problem Let n N. ßÞ Prove the Ðinequality 1 n n n+1 > Problem Prove that Õ π 2 = 2cos 2 n+1 n radical signs for n N. 22 Problem Let a 1 = 3,b 1 = 4, and a n = 3 a n 1,b n = 4 b n 1 when n > 1. Prove that a 1000 > b Problem Let n N,n > 1. Prove that (2n 1) (2n) < 1 3n Problem Prove that if n is a natural number, then n (3n 1) = n 2 (n+1). 25 Problem Prove that if n is a natural number, then (2n 1) 2 = n(4n2 1) Problem Prove that for all natural numbers n > 1. 4 n n+1 < (2n)! (n!) 2 27 Problem Prove that the sum of the cubes of three consecutive positive integers is divisible by 9.

12 8 Chapter 1 28 Problem If x 1,n N prove that equals 1 1+x x x n x8 1+x 2n 1 x 1 + 2n+1 1 x 2n Problem Is it true that for every natural number n the quantity n 2 + n+41 is a prime? Prove or disprove! 30 Problem Give an example of an assertion which is not true for any positive integer, yet for which the induction step holds. 31 Problem Give an example of an assertion which is true for the first two million positive integers but fails for every integer greater than Problem Prove by induction on n that a set having n elements has exactly 2 n subsets. 33 Problem Prove that if n is a natural number, is always an integer. n 5 /5+n 4 /2+n 3 /3 n/30 34 Problem (Halmos) ) Every man in a village knows instantly when another s wife is unfaithful, but never when his own is. Each man is completely intelligent and knows that every other man is. The law of the village demands that when a man can PROVE that his wife has been unfaithful, he must shoot her before sundown the same day. Every man is completely law-abiding. One day the mayor announces that there is at least one unfaithful wife in the village. The mayor always tells the truth, and every man believes him. If in fact there are exactly forty unfaithful wives in the village (but that fact is not known to the men,) what will happen after the mayor s announcement? 35 Problem 1. Let a 1,a 2,...a n be positive real numbers with a 1 a 2 a n = 1. Use induction to prove that a 1 + a 2 + +a n n, with equality if and only if a 1 = a 2 = = a n = Use the preceding part to give another proof of the Arithmetic-Mean-Geometric-Mean Inequality. 3. Prove that if n > 1, then 4. Prove that if n > 1 then (2n 1) < n n. n (n+1) 1/n 1 < n. 5. Prove that if n > 1 then n < n 1 1 (n+1) 1/n + 1 n Given that u, v, w are positive, 0 < a 1, and that u+v+w = 1, prove that 1 u a 1 v a 1 w a 27 27a+9a 2 a Let y 1,y 2,...,y n be positive real numbers. Prove the Harmonic-Mean- Geometric-Mean Inequality: n n y1 y 2 y n. y 1 y 2 y n 8. Let a 1,...,a n be positive real numbers, all different. Set s = a 1 + a 2 + +a n. (a) Prove that (n 1) s < 1 1. s a 1 r n r a 1 r n r (b) Deduce that 4n s < 1 a r (s a r ) n 1 < n 1. a r 1 r n 1 r n 36 Problem Suppose that x 1,x 2,...,x n are nonnegative real numbers with x 1 + x 2 + +x n 1/2. Prove that (1 x 1 )(1 x 2 ) (1 x n ) 1/2. 37 Problem Given a positive integer n prove that there is a polynomial T n such that cosnx = T n (cosx) for all real numbers x. T n is called the n-th Tchebychev Polynomial. 38 Problem Prove that for all natural numbers n > 1. 1 n n n > 13 24

13 Fibonacci Numbers 9 39 Problem In how many regions will a sphere be divided by n planes passing through its centre if no three planes pass through one and the same diameter? 40 Problem (IMO 1977) Let f, f : N N be a function satisfying f(n+1) > f(f(n)) for each positive integer n. Prove that f(n) = n for each n. 41 Problem Let F 0 (x) = x,f(x) = 4x(1 x),f n+1 (x) = F(F n (x)),n = 0,1,... Prove that 1 (Hint: Let x = sin 2 θ.) 0 F n (x)dx = 22n 1 2 2n Fibonacci Numbers The Fibonacci numbers f n are given by the recurrence f 0 = 0, f 1 = 1, f n+1 = f n 1 + f n, n 1. (1.5) Thus the first few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21,.... A number of interesting algebraic identities can be proved using the above recursion. 42 Example Prove that Solution: We have Summing both columns, as desired. 43 Example Prove that f 1 + f f n = f n+2 1. f 1 = f 3 f 2 f 2 = f 4 f 3 f 3 = f 5 f 4. f n = f n+2 f n+1. f 1 + f f n = f n+2 f 2 = f n+2 1, f 1 + f 3 + f f 2n 1 = f 2n. Solution: Observe that Adding columnwise we obtain the desired identity. f 1 = f 2 f 0 f 3 = f 4 f 2 f 5 = f 6 f 4... f 2n 1 = f 2n f 2n 2 44 Example Prove that Solution: We have Thus f f f 2 n = f n f n+1. f n 1 f n+1 = ( f n+1 f n )( f n + f n 1 ) = f n+1 f n f 2 n + f n+1 f n 1 f n f n 1. f n+1 f n f n f n 1 = f 2 n,

14 10 Chapter 1 which yields f f f 2 n = f n f n Theorem (Cassini s Identity) f n 1 f n+1 f 2 n = ( 1) n, n 1. Proof: Observe that f n 1 f n+1 fn 2 = ( f n f n 2 )( f n + f n 1 ) fn 2 = f n 2 f n f n 1 ( f n 2 f n ) = ( f n 2 f n fn 1) 2 Thus if v n = f n 1 f n+1 f 2 n, we have v n = v n 1. This yields v n = ( 1) n 1 v 1 which is to say 46 Example (IMO 1981) Determine the maximum value of f n 1 f n+1 f 2 n = ( 1) n 1 ( f 0 f 2 f 2 1 ) = ( 1) n. m 2 + n 2, where m, n are positive integers satisfying m, n {1, 2, 3,..., 1981} and (n 2 mn m 2 ) 2 = 1. Solution: Call a pair (n,m) admissible if m,n {1,2,...,1981} and (n 2 mn m 2 ) 2 = 1. If m = 1, then (1,1) and (2,1) are the only admissible pairs. Suppose now that the pair (n 1,n 2 ) is admissible, with n 2 > 1. As n 1 (n 1 n 2 ) = n 2 2 ± 1 > 0, we must have n 1 > n 2. Let now n 3 = n 1 n 2. Then 1 = (n 2 1 n 1 n 2 n 2 2) 2 = (n 2 2 n 2 n 3 n 2 3) 2, making (n 2,n 3 ) also admissible. If n 3 > 1, in the same way we conclude that n 2 > n 3 and we can let n 4 = n 2 n 3 making (n 3,n 4 ) an admissible pair. We have a sequence of positive integers n 1 > n 2 >..., which must necessarily terminate. This terminates when n k = 1 for some k. Since (n k 1,1) is admissible, we must have n k 1 = 2. The sequence goes thus 1,2,3,5,8,...,987,1597, i.e., a truncated Fibonacci sequence. The largest admissible pair is thus (1597, 987) and so the maximum sought is Let τ = be the Golden Ratio. Observe that τ 1 =. The number τ is a root of the quadratic equation 2 2 x 2 = x+1. We now obtain a closed formula for f n. We need the following lemma. 47 Lemma If x 2 = x+1,n 2 then we have x n = f n x+ f n 1. Proof: We prove this by induction on n. For n = 2 the assertion is a triviality. Assume that n > 2 and that x n 1 = f n 1 x+ f n 2. Then x n = x n 1 x = ( f n 1 x+ f n 2 )x = f n 1 (x+1)+ f n 2 x = ( f n 1 + f n 2 )x+ f n 1 = f n x+ f n 1 48 Theorem (Binet s Formula) The n-th Fibonacci number is given by n = 0,2,... f n = n n

15 Practice 11 Proof: The roots of the equation x 2 = x+1 are τ = τ n = τ f n + f n 1 and 1 τ = 1 5. In virtue of the above lemma, 2 and Subtracting from where Binet s Formula follows. (1 τ) n = (1 τ) f n + f n 1. τ n (1 τ) n = 5 f n, 49 Example (Cesàro) Prove that Solution: Using Binet s Formula, n k=0 n k 2 k f k = = = n k=0 n k 2 k f k = f 3n. n k=0 n k 2 k τk (1 τ) k 5 n k=0 n 1 5 n k=0 n k τ k 1 ((1+2τ) n (1+2(1 τ)) n ). 5 k 2 k (1 τ) k As τ 2 = τ + 1,1+2τ = τ 3. Similarly 1+2(1 τ) = (1 τ) 3. Thus as wanted. The following theorem will be used later. n k=0 n k 2 k f k = 1 5 (τ) 3n +(1 τ) 3n = f 3n, 50 Theorem If s 1, t 0 are integers then f s+t = f s 1 f t + f s f t+1. Proof: We keep t fixed and prove this by using strong induction on s. For s = 1 we are asking whether f t+1 = f 0 f t + f 1 f t+1, which is trivially true. Assume that s > 1 and that f s k+t = f s k 1 f t + f s k f t+1 for all k satisfying 1 k s 1. We have f s+t = f s+t 1 + f s+t 2 by the Fibonacci recursion, = f s 1+t + f s 2+t trivially, = f s 2 f t + f s 1 f t+1 + f s 3 f t + f s 2 f t+1 by the inductive assumption = f t ( f s 2 + f s 3 )+ f t+1 ( f s 1 + f s 2 ) rearranging, = f t f s 1 + f t+1 f s by the Fibonacci recursion. This finishes the proof. Practice

16 12 Chapter 1 51 Problem Prove that 52 Problem Prove that 53 Problem Prove that f n+1 f n f n 1 f n 2 = f 2n 1, n > 2. f 2 n+1 = 4 f n f n 1 + f 2 n 2, n > 1. f 1 f 2 + f 2 f f 2n 1 f 2n = f 2 2n. 54 Problem Let N be a natural number. Prove that the largest n such that f n N is given by log N n = log Problem Prove that f 2 n + f 2 n 1 = f 2n Problem Prove that 62 Problem Prove that 63 Problem Prove that 64 Problem Prove that 1 arctan = π/4. f n=1 2n+1 f n lim n τ n = 1. 5 f n+r lim = τ r. n f n 1 f n k=0 2 k = 2+ f 2 n 2. f 2 n 56 Problem Prove that if n > 1, f 2 n f n+l f n l = ( 1) n+l f 2 l. Deduce that k=0 1 f 2 k = Problem Prove that n k=1 58 Problem Prove that f 2k = n=2 (n k) f 2k+1. n k=0 1 f n 1 f n+1 = Problem (Cesàro) Prove that 66 Problem Prove that n k=0 n k f k = f 2n. Hint: What is 1 f n 1 f n 1 f n f n+1? is a rational number. n=1 f n 10 n 59 Problem Prove that 60 Problem Prove that n=1 f n f n+1 f n+2 = 1. 1/ f 2 n = 4 τ. n=0 67 Problem Find the exact value of 1994 k=1( 1) k 1995 k. k f 68 Problem Prove the converse of Cassini s Identity: If k and m are integers such that m 2 km k 2 = 1, then there is an integer n such that k = ± f n,m = ± f n+1.

17 Pigeonhole Principle Pigeonhole Principle The Pigeonhole Principle states that if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons. This apparently trivial principle is very powerful. Let us see some examples. 69 Example (Putnam 1978) Let A be any set of twenty integers chosen from the arithmetic progression 1,4,...,100. Prove that there must be two distinct integers in A whose sum is 104. Solution: We partition the thirty four elements of this progression into nineteen groups {1},{52}, {4,100}, {7,97}, {10,94},... {49,55}. Since we are choosing twenty integers and we have nineteen sets, by the Pigeonhole Principle there must be two integers that belong to one of the pairs, which add to Example Show that amongst any seven distinct positive integers not exceeding 126, one can find two of them, say a and b, which satisfy b < a 2b. Solution: Split the numbers {1, 2, 3,..., 126} into the six sets {1,2},{3,4,5,6},{7,8,...,13,14},{15,16,...,29,30}, {31,32,...,61,62} and {63,64,...,126}. By the Pigeonhole Principle, two of the seven numbers must lie in one of the six sets, and obviously, any such two will satisfy the stated inequality. 71 Example Given any set of ten natural numbers between 1 and 99 inclusive, prove that there are two disjoint nonempty subsets of the set with equal sums of their elements. Solution: There are = 1023 non-empty subsets that one can form with a given 10-element set. To each of these subsets we associate the sum of its elements. The maximum value that any such sum can achieve is = 945 < Therefore, there must be at least two different subsets that have the same sum. 72 Example No matter which fifty five integers may be selected from prove that one must select some two that differ by 10. {1,2,...,100}, Solution: First observe that if we choose n+1 integers from any string of 2n consecutive integers, there will always be some two that differ by n. This is because we can pair the 2n consecutive integers into the n pairs {a+1,a+2,a+3,...,a+2n} {a+1,a+n+1},{a+2,a+n+2},...,{a+n,a+2n}, and if n+1 integers are chosen from this, there must be two that belong to the same group. So now group the one hundred integers as follows: and {1,2,...20},{21,22,...,40}, {41,42,...,60}, {61,62,...,80} {81,82,...,100}. If we select fifty five integers, we must perforce choose eleven from some group. From that group, by the above observation (let n = 10), there must be two that differ by 10.

18 14 Chapter 1 73 Example (AHSME 1994) Label one disc 1, two discs 2, three discs 3,..., fifty discs 50. Put these = 1275 labeled discs in a box. Discs are then drawn from the box at random without replacement. What is the minimum number of discs that must me drawn in order to guarantee drawing at least ten discs with the same label? Solution: If we draw all the = 45 labelled 1,..., 9 and any nine from each of the discs 10,..., 50, we have drawn = 414 discs. The 415-th disc drawn will assure at least ten discs from a label. 74 Example (IMO 1964) Seventeen people correspond by mail with one another each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there at least three people who write to each other about the same topic. Solution: Choose a particular person of the group, say Charlie. He corresponds with sixteen others. By the Pigeonhole Principle, Charlie must write to at least six of the people of one topic, say topic I. If any pair of these six people corresponds on topic I, then Charlie and this pair do the trick, and we are done. Otherwise, these six correspond amongst themselves only on topics II or III. Choose a particular person from this group of six, say Eric. By the Pigeonhole Principle, there must be three of the five remaining that correspond with Eric in one of the topics, say topic II. If amongst these three there is a pair that corresponds with each other on topic II, then Eric and this pair correspond on topic II, and we are done. Otherwise, these three people only correspond with one another on topic III, and we are done again. 75 Example Given any seven distinct real numbers x 1,...x 7, prove that we can always find two, say a,b with 0 < a b 1+ab < 1 3. Solution: Put x k = tana k for a k satisfying π 2 < a k < π 2. Divide the interval ( π 2, π ) into six non-overlapping subintervals of 2 equal length. By the Pigeonhole Principle, two of seven points will lie on the same interval, say a i < a j. Then 0 < a j a i < π 6. Since the tangent increases in ( π/2,π/2), we obtain as desired. 0 < tan(a j a i ) = tana j tana i 1+tana j tana i < tan π 6 = 1 3, 76 Example (Canadian Math Olympiad 1981) Let a 1,a 2,...,a 7 be nonnegative real numbers with If a 1 + a a 7 = 1. M = max 1 k 5 a k + a k+1 + a k+2, determine the minimum possible value that M can take as the a k vary. Solution: Since a 1 a 1 + a 2 a 1 + a 2 + a 3 and a 7 a 6 + a 7 a 5 + a 6 + a 7 we see that M also equals max {a 1,a 7,a 1 + a 2,a 6 + a 7,a k + a k+1 + a k+2 }. 1 k 5 We are thus taking the maximum over nine quantities that sum 3(a 1 + a 2 + +a 7 ) = 3. These nine quantities then average 3/9 = 1/3. By the Pigeonhole Principle, one of these is 1/3, i.e. M 1/3. If a 1 = a 1 + a 2 = a 1 + a 2 + a 3 = a 2 + a 3 + a 4 = a 3 +a 4 +a 5 = a 4 +a 5 +a 6 = a 5 +a 6 +a 7 = a 7 = 1/3, we obtain the 7-tuple (a 1,a 2,a 3,a 4,a 5,a 6,a 7 ) = (1/3,0,0,1/3,0,0,1/3), which shows that M = 1/3. Practice

19 Practice Problem (AHSME 1991) A circular table has exactly sixty chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest possible value of N? Answer: Problem Show that if any five points are all in, or on, a square of side 1, then some pair of them will be at most at distance 2/2. 79 Problem (Eötvös, 1947) Prove that amongst six people in a room there are at least three who know one another, or at least three who do not know one another. 80 Problem Show that in any sum of non-negative real numbers there is always one number which is at least the average of the numbers and that there is always one member that it is at most the average of the numbers. 81 Problem We call a set sum free if no two elements of the set add up to a third element of the set. What is the maximum size of a sum free subset of {1,2,...,2n 1}. Hint: Observe that the set {n+1,n+2,...,2n 1} of n+1 elements is sum free. Show that any subset with n+2 elements is not sum free. 82 Problem (MMPC 1992) Suppose that the letters of the English alphabet are listed in an arbitrary order. 1. Prove that there must be four consecutive consonants. 2. Give a list to show that there need not be five consecutive consonants. 3. Suppose that all the letters are arranged in a circle. Prove that there must be five consecutive consonants. 83 Problem (Stanford 1953) Bob has ten pockets and forty four silver dollars. He wants to put his dollars into his pockets so distributed that each pocket contains a different number of dollars. 1. Can he do so? 2. Generalise the problem, considering p pockets and n dollars. The problem is most interesting when Why? n = (p 1)(p 2) Problem No matter which fifty five integers may be selected from {1,2,...,100}, prove that you must select some two that differ by 9, some two that differ by 10, some two that differ by 12, and some two that differ by 13, but that you need not have any two that differ by Problem Let mn + 1 different real numbers be given. Prove that there is either an increasing sequence with at least n+1 members, or a decreasing sequence with at least m+1 members. 86 Problem If the points of the plane are coloured with three colours, show that there will always exist two points of the same colour which are one unit apart. 87 Problem Show that if the points of the plane are coloured with two colours, there will always exist an equilateral triangle with all its vertices of the same colour. There is, however, a colouring of the points of the plane with two colours for which no equilateral triangle of side 1 has all its vertices of the same colour. 88 Problem Let r 1,r 2,...,r n,n > 1 be real numbers of absolute value not exceeding 1 and whose sum is 0. Show that there is a non-empty proper subset whose sum is not more than 2/n in size. Give an example in which any subsum has absolute 1 value at least n Problem Let r 1,r 2,...,r n be real numbers in the interval [0,1]. Show that there are numbers ε k,1 k n,ε k = 1,0,1 not all zero, such ε k r k that n k=1 n 2 n. 90 Problem (USAMO, 1979) Nine mathematicians meet at an international conference and discover that amongst any three of them, at least two speak a common language. If each of the mathematicians can speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language. 91 Problem (USAMO, 1982) In a party with 1982 persons, amongst any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?

20 16 Chapter 1 92 Problem (USAMO, 1985) There are n people at a party. Prove that there are two people such that, of the remaining n 2 people, there are at least n/2 1 of them, each of whom knows both or else knows neither of the two. Assume that knowing is a symmetrical relationship. 93 Problem (USAMO, 1986) During a certain lecture, each of five mathematicians fell asleep exactly twice. For each pair of these mathematicians, there was some moment when both were sleeping simultaneously. Prove that, at some moment, some three were sleeping simultaneously. 94 Problem Let P n be a set of en! +1 points on the plane. Any two distinct points of P n are joined by a straight line segment which is then coloured in one of n given colours. Show that at least one monochromatic triangle is formed. (Hint: e = 1/n!.) n=0

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### Stanford University Educational Program for Gifted Youth (EPGY) Number Theory. Dana Paquin, Ph.D.

Stanford University Educational Program for Gifted Youth (EPGY) Dana Paquin, Ph.D. paquin@math.stanford.edu Summer 2010 Note: These lecture notes are adapted from the following sources: 1. Ivan Niven,

### Homework until Test #2

MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.

Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole

### San Jose Math Circle October 17, 2009 ARITHMETIC AND GEOMETRIC PROGRESSIONS

San Jose Math Circle October 17, 2009 ARITHMETIC AND GEOMETRIC PROGRESSIONS DEFINITION. An arithmetic progression is a (finite or infinite) sequence of numbers with the property that the difference between

### If a question asks you to find all or list all and you think there are none, write None.

If a question asks you to find all or list all and you think there are none, write None 1 Simplify 1/( 1 3 1 4 ) 2 The price of an item increases by 10% and then by another 10% What is the overall price

### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion

CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a

### Prime Numbers. Chapter Primes and Composites

Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12

CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.

### Proof: A logical argument establishing the truth of the theorem given the truth of the axioms and any previously proven theorems.

Math 232 - Discrete Math 2.1 Direct Proofs and Counterexamples Notes Axiom: Proposition that is assumed to be true. Proof: A logical argument establishing the truth of the theorem given the truth of the

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### Review for Final Exam

Review for Final Exam Note: Warning, this is probably not exhaustive and probably does contain typos (which I d like to hear about), but represents a review of most of the material covered in Chapters

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

### Appendix F: Mathematical Induction

Appendix F: Mathematical Induction Introduction In this appendix, you will study a form of mathematical proof called mathematical induction. To see the logical need for mathematical induction, take another

### Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

### Number Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures

Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it

### 1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3.

1 By how much does 1 3 of 5 exceed 1 of 1 3? What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3 A ticket fee was \$10, but then it was reduced The number of customers

### NUMBER THEORY AMIN WITNO

NUMBER THEORY AMIN WITNO ii Number Theory Amin Witno Department of Basic Sciences Philadelphia University JORDAN 19392 Originally written for Math 313 students at Philadelphia University in Jordan, this

### Continued fractions and good approximations.

Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our

### PEN A Problems. a 2 +b 2 ab+1. 0 < a 2 +b 2 abc c, x 2 +y 2 +1 xy

Divisibility Theory 1 Show that if x,y,z are positive integers, then (xy + 1)(yz + 1)(zx + 1) is a perfect square if and only if xy +1, yz +1, zx+1 are all perfect squares. 2 Find infinitely many triples

### SUM OF TWO SQUARES JAHNAVI BHASKAR

SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted

### Algebraic Systems, Fall 2013, September 1, 2013 Edition. Todd Cochrane

Algebraic Systems, Fall 2013, September 1, 2013 Edition Todd Cochrane Contents Notation 5 Chapter 0. Axioms for the set of Integers Z. 7 Chapter 1. Algebraic Properties of the Integers 9 1.1. Background

### SECTION 10-2 Mathematical Induction

73 0 Sequences and Series 6. Approximate e 0. using the first five terms of the series. Compare this approximation with your calculator evaluation of e 0.. 6. Approximate e 0.5 using the first five terms

### MATH REVIEW KIT. Reproduced with permission of the Certified General Accountant Association of Canada.

MATH REVIEW KIT Reproduced with permission of the Certified General Accountant Association of Canada. Copyright 00 by the Certified General Accountant Association of Canada and the UBC Real Estate Division.

### Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

### Lecture 3. Mathematical Induction

Lecture 3 Mathematical Induction Induction is a fundamental reasoning process in which general conclusion is based on particular cases It contrasts with deduction, the reasoning process in which conclusion

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

Mathematics 0N1 Solutions 1 1. Write the following sets in list form. 1(i) The set of letters in the word banana. {a, b, n}. 1(ii) {x : x 2 + 3x 10 = 0}. 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi)

### Arithmetic and Geometric Sequences

Arithmetic and Geometric Sequences Felix Lazebnik This collection of problems is for those who wish to learn about arithmetic and geometric sequences, or to those who wish to improve their understanding

### An Introductory Course in Elementary Number Theory. Wissam Raji

An Introductory Course in Elementary Number Theory Wissam Raji 2 Preface These notes serve as course notes for an undergraduate course in number theory. Most if not all universities worldwide offer introductory

### vertex, 369 disjoint pairwise, 395 disjoint sets, 236 disjunction, 33, 36 distributive laws

Index absolute value, 135 141 additive identity, 254 additive inverse, 254 aleph, 466 algebra of sets, 245, 278 antisymmetric relation, 387 arcsine function, 349 arithmetic sequence, 208 arrow diagram,

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### Section 6-2 Mathematical Induction

6- Mathematical Induction 457 In calculus, it can be shown that e x k0 x k k! x x x3!! 3!... xn n! where the larger n is, the better the approximation. Problems 6 and 6 refer to this series. Note that

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University

Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### Taylor and Maclaurin Series

Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions

### Introduction. Appendix D Mathematical Induction D1

Appendix D Mathematical Induction D D Mathematical Induction Use mathematical induction to prove a formula. Find a sum of powers of integers. Find a formula for a finite sum. Use finite differences to

### It is time to prove some theorems. There are various strategies for doing

CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it

### Handout NUMBER THEORY

Handout of NUMBER THEORY by Kus Prihantoso Krisnawan MATHEMATICS DEPARTMENT FACULTY OF MATHEMATICS AND NATURAL SCIENCES YOGYAKARTA STATE UNIVERSITY 2012 Contents Contents i 1 Some Preliminary Considerations

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Chapter 2 Limits Functions and Sequences sequence sequence Example

Chapter Limits In the net few chapters we shall investigate several concepts from calculus, all of which are based on the notion of a limit. In the normal sequence of mathematics courses that students

### PROBLEM SET 7: PIGEON HOLE PRINCIPLE

PROBLEM SET 7: PIGEON HOLE PRINCIPLE The pigeonhole principle is the following observation: Theorem. Suppose that > kn marbles are distributed over n jars, then one jar will contain at least k + marbles.

### CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS

CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas

### INTRODUCTORY SET THEORY

M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,

### TRIANGLES ON THE LATTICE OF INTEGERS. Department of Mathematics Rowan University Glassboro, NJ Andrew Roibal and Abdulkadir Hassen

TRIANGLES ON THE LATTICE OF INTEGERS Andrew Roibal and Abdulkadir Hassen Department of Mathematics Rowan University Glassboro, NJ 08028 I. Introduction In this article we will be studying triangles whose

### Lecture 1: Elementary Number Theory

Lecture 1: Elementary Number Theory The integers are the simplest and most fundamental objects in discrete mathematics. All calculations by computers are based on the arithmetical operations with integers

### An example of a computable

An example of a computable absolutely normal number Verónica Becher Santiago Figueira Abstract The first example of an absolutely normal number was given by Sierpinski in 96, twenty years before the concept

### Chapter 7. Induction and Recursion. Part 1. Mathematical Induction

Chapter 7. Induction and Recursion Part 1. Mathematical Induction The principle of mathematical induction is this: to establish an infinite sequence of propositions P 1, P 2, P 3,..., P n,... (or, simply

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### 1. R In this and the next section we are going to study the properties of sequences of real numbers.

+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

### Chapter Two. Number Theory

Chapter Two Number Theory 2.1 INTRODUCTION Number theory is that area of mathematics dealing with the properties of the integers under the ordinary operations of addition, subtraction, multiplication and

### Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions

A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25

### Mathematical Induction

Chapter 4 Mathematical Induction 4.1 The Principle of Mathematical Induction Preview Activity 1 (Exploring Statements of the Form.8n N/.P.n//) One of the most fundamental sets in mathematics is the set

### CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### 1. Determine all real numbers a, b, c, d that satisfy the following system of equations.

altic Way 1999 Reykjavík, November 6, 1999 Problems 1. etermine all real numbers a, b, c, d that satisfy the following system of equations. abc + ab + bc + ca + a + b + c = 1 bcd + bc + cd + db + b + c

### Just the Factors, Ma am

1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive

3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### Problem Set 5. AABB = 11k = (10 + 1)k = (10 + 1)XY Z = XY Z0 + XY Z XYZ0 + XYZ AABB

Problem Set 5 1. (a) Four-digit number S = aabb is a square. Find it; (hint: 11 is a factor of S) (b) If n is a sum of two square, so is 2n. (Frank) Solution: (a) Since (A + B) (A + B) = 0, and 11 0, 11

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### 2 The Euclidean algorithm

2 The Euclidean algorithm Do you understand the number 5? 6? 7? At some point our level of comfort with individual numbers goes down as the numbers get large For some it may be at 43, for others, 4 In

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

### Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

### Mathematical Induction

Mathematical Induction Victor Adamchik Fall of 2005 Lecture 2 (out of three) Plan 1. Strong Induction 2. Faulty Inductions 3. Induction and the Least Element Principal Strong Induction Fibonacci Numbers

### 2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2

00 Problem If a and b are nonzero real numbers such that a b, compute the value of the expression ( ) ( b a + a a + b b b a + b a ) ( + ) a b b a + b a +. b a a b Answer: 8. Solution: Let s simplify the

### Mathematics of Cryptography

CHAPTER 2 Mathematics of Cryptography Part I: Modular Arithmetic, Congruence, and Matrices Objectives This chapter is intended to prepare the reader for the next few chapters in cryptography. The chapter

### Solutions to Practice Problems

Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2

### 3. Recurrence Recursive Definitions. To construct a recursively defined function:

3. RECURRENCE 10 3. Recurrence 3.1. Recursive Definitions. To construct a recursively defined function: 1. Initial Condition(s) (or basis): Prescribe initial value(s) of the function.. Recursion: Use a

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Baltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions

Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.

### MATH10040 Chapter 2: Prime and relatively prime numbers

MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive

### THE CONGRUENT NUMBER PROBLEM

THE CONGRUENT NUMBER PROBLEM KEITH CONRAD 1. Introduction A right triangle is called rational when its legs and hypotenuse are all rational numbers. Examples of rational right triangles include Pythagorean

### 2. Methods of Proof Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try.

2. METHODS OF PROOF 69 2. Methods of Proof 2.1. Types of Proofs. Suppose we wish to prove an implication p q. Here are some strategies we have available to try. Trivial Proof: If we know q is true then

### MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions

### Mathematical induction & Recursion

CS 441 Discrete Mathematics for CS Lecture 15 Mathematical induction & Recursion Milos Hauskrecht milos@cs.pitt.edu 5329 Sennott Square Proofs Basic proof methods: Direct, Indirect, Contradiction, By Cases,

### New Zealand Mathematical Olympiad Committee. Induction

New Zealand Mathematical Olympiad Committee Induction Chris Tuffley 1 Proof by dominoes Mathematical induction is a useful tool for proving statements like P is true for all natural numbers n, for example

### PROOFS BY DESCENT KEITH CONRAD

PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Continued Fractions. Darren C. Collins

Continued Fractions Darren C Collins Abstract In this paper, we discuss continued fractions First, we discuss the definition and notation Second, we discuss the development of the subject throughout history

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE

CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE With the notion of bijection at hand, it is easy to formalize the idea that two finite sets have the same number of elements: we just need to verify

### 8 Divisibility and prime numbers

8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express