On the greatest and least prime factors of n! + 1, II


 Morgan Richard
 2 years ago
 Views:
Transcription
1 Publ. Math. Debrecen Manuscript May 7, 004 On the greatest and least prime factors of n! + 1, II By C.L. Stewart In memory of Béla Brindza Abstract. Let ε be a positive real number. We prove that for infinitely many odd integers n the least prime factor of n! + 1 is at most 8 + ε n and that for infinitely many positive integers n the greatest prime factor of n!+1 exceeds 11 εn. 1. Introduction In 1856, Liouville [6] proved that p 1! + 1 is not a power of p for any prime p larger than 5. More than a century later Erdős and Graham [] asked if the equation p 1! + a p 1 = p k 1 has only finitely many solutions in positive integers a,k,p with p an odd prime. In 1991 Brindza and Erdős [1] resolved the question by proving that all solutions of 1 are smaller than an effectively computable number. A few years later Yu and Liu [10] and also Le [5] determined the complete list of solutions. Mathematics Subject Classification: 11D75,11J5. Key words and phrases: shifted factorials, prime factors. This research was supported in part by Grant A358 from the Natural Sciences and Engineering Research Council of Canada and by the Canada Research Chairs Program.
2 C.L. Stewart In 1976 we investigated with Erdős [3] the arithmetical character of integers of the form n! + 1 where n is a positive integer. For any integer m larger than 1 let Pm denote the greatest prime factor of m and let pm denote the least prime factor of m. By Wilson s theorem p divides p 1! + 1 whenever p is a prime. Since all prime factors of n! + 1 exceed n we see that pn! + 1 = n + 1 whenever n + 1 is a prime. We showed with Erdős [3] that if n + 1 is not a prime then Further, for almost all integers n, log n pn! + 1 > n + 1 o1 log log n. pn! + 1 > n + εnn 1, 3 where εn is any positive function that decreases to 0 as n. In [3] we indicated how to prove that for infinitely many integers n for which n + 1 is not a prime pn! + 1 is less than n. We observed, as a direct consequence of Wilson s theorem, that if p is a prime then p 1 i!i! 1 i+1 mod p, 0 i p 1. 4 Thus if p i!+1 for some odd integer i > 1 then, from 4, p p i 1!+1. For any positive real numbers θ and t with t > 1 θ we have and Note that 5 and 6 that p p i 1 = 1 p 1 p i p max p i, max θ, min θ, 1 t θ 1 t 5 1 t θ 1 t. 6 and so on taking θ = p i p p i 1 and t = p 1 p p p 1 = + p 1, we see from and p min i, p + p i 1 p 1,
3 On the greatest and least prime factors of n! + 1, II 3 for 0 < i < p 1. As i tends to infinity so also do p and p i. Thus for each ε > 0, pn! + 1 < + εn, 7 for infinitely many composite integers n + 1. Further Pn! + 1 > n, 8 for infinitely many positive integers n. We indicated in [3] that +ε in 7 could be replaced by δ for some positive number δ. Our first result will be of this character. Theorem 1. Let ε > 0. There are infinitely many odd integers n for which pn! + 1 < + ε n. 9 8 Observe that = With Erdős [3] we proved that holds with Pn! + 1 in place of pn! + 1 for all positive integers n. Of course this only is an improvement on for those integers n for which n + 1 is a prime. Recently Luca and Shparlinski [7] sharpened this result by proving that 1 Pn! + 1 > n o1 log n; 10 indeed they established 10 with Pn!+1 replaced by Pn!+fn where f is any nonzero polynomial with integer coefficients. In 00 Murty and Wong [8] showed that if the abc conjecture holds, then Pn! + 1 > 1 + o1n log n. In 1976 we improved on 8 with Erdős [3] by proving that there is a positive number δ such that Pn! + 1 > + δn, 11 for infinitely many integers n. Luca and Shparlinski [7] established 11 with +δn replaced by 5 + o1 n and showed that their result applies with n! + fn in place of n! + 1 where f is any nonzero polynomial with integer coefficients. Our next result gives a further improvement on 11.
4 4 C.L. Stewart For any set X we denote the cardinality of X by X. For any set A of positive integers and any positive integer n we put An = A {1,...,n}. The lower asymptotic density of A is lim inf An n. Theorem. Let ε > 0. The set of positive integers n for which 11 Pn! + 1 > ε n 1 has positive lower asymptotic density. As we remarked in [3] estimates, 3, 7, 8 and 11 hold with n! + 1 replaced by n! 1 and the same comment applies to the estimates 9 and 1. Further, the same techniques used to prove Theorems 1 and allow one to prove, for instance, that for each positive real number ε there exist infinitely many positive integers n for which Pn! + 1 > ε n and there exist infinitely many positive integers n for which Pn! + 1n! 1 > ε n.. Preliminary lemmas Let p 1,p,... denote the sequence of prime numbers and put d k = p k+1 p k for k = 1,,.... Our first lemma, due to HeathBrown, gives a bound on the frequency of large differences between consecutive prime numbers. Lemma 1. Let ε be a positive real number. There is a positive number c, which depends on ε, such that p k x d k < cx3 18 +ε. Proof. This is Theorem 1 of [4]. Our next result gives a bound for the size of the greatest common divisor of a collection of terms of the form k! + 1.
5 On the greatest and least prime factors of n! + 1, II 5 Lemma. Let n and t be positive integers with t and let i 1,...,i t be distinct positive integers from a subinterval of [1,n] of length l. Then gcdi 1! + 1,...,i t! + 1 < n l t Further, there exists a positive number c 1 such that if n exceeds c 1 and t 3 then gcdi 1! + 1,...,i t! + 1 < e n n l t Furthermore, let δ and ε be positive real numbers. There exists a positive number c, which depends on ε and δ, such that if n exceeds c, and then gcdi 1! + 1,...,i t! + 1 < exp 1 + εl log t t 3 t < n δ, 15 n l >, 16 log n 1 + log ne l + t log n max1,log log t t We remark that it is possible to replace the term max1,log log t on the right hand side of inequality 17 by ft where f is any real valued function to the real numbers of size at least 1 for which lim ft = t provided that c is modified to depend on f. Proof of Lemma. Let A be a positive integer and let k 1,k and k 3,k 4 be distinct pairs of integers with n k 1 > k 1 and n k 3 > k 4 1 for which A k i! + 1, 18 for i = 1,, 3, 4. Suppose, without loss of generality, that k 1 k k 3 k Note that {k 1,k,k 3,k 4 } may only contain 3 elements. Then, by 18, A k i! k i+1!, for i = 1,3,
6 6 C.L. Stewart and so A k i+1!k i k i 1 k i , for i = 1,3. But, by 18, gcda,k i+1! = 1 for i = 1,3 hence for i = 1, 3. Therefore Since A k i k i , 0 A k 1 k + 1 k 3 k k 1 k + 1 k 3 k = k 3 k 4! k1 k! k 3 k 4! gcda,k 3 k 4! = 1 and k 1 k k 3 k 4, we find that k1 k k3 k 4, A k 3 k 4! 1 k 1 k + 1 k 3 k Notice that k 1 k + 1 k 3 k < n k 1 k. Thus, provided that k 1 k + 1 k 3 k 4 + 1, 3 we deduce from 1, and the fact that m! m m e, when m is a positive integer, that A < n k 1 k k 3 k 4 ne k3 k 4. 4 k 3 k 4 Since gx = ne x x attains its maximum value at x = n we see that, when 3 holds, A < n k 1 k k 3 k 4 e n. 5 We now put A = gcdi 1! + 1,...,i t! + 1. We shall prove 13 first. Put µ = min{i a i b i a > i b, a,b {1,...,t}}
7 On the greatest and least prime factors of n! + 1, II 7 and let k 1,k be elements of {i 1,...,i t } with k 1 k = µ. Since µ we see from 0 with i = 1 that A < n k 1 k n l t 1, l t 1 as required. We shall prove 14 next. There are t pairs of integers i,i with i > i which can be chosen from {i 1,...,i t }. Associated to each such pair is the difference i i and 0 < i i l. Therefore there are two such pairs, k 1,k and k 3,k 4 say, for which 0 k 1 k k 3 k 4 Thus, provided that 3 holds, by 5 and 6, A < n 1 t e n, hence, since t 3 and t 1 t 1, A < e n n l t 1. l l t 1. 6 It remains only to ensure that 3 holds. We may assume that A exceeds e n since otherwise the result is immediate. Note that if k 1 = k 3 then, since the pairs k 1,k and k 3,k 4 are distinct, k k 4 and so 3 holds. Thus we may assume that k 1 > k 3 ; a similar argument applies if k 3 < k 1. Further, we may assume, after renumbering k,k 3,k 4 if necessary, that k k 3 > k 4. Since, as in 0, A divides k 1 k we see that A n k 1 k. But A exceeds e n and so k 1 k n log n. By a version of the prime number theorem with an explicit error term, for n sufficiently large there is a prime p between k 1 and k. As a consequence p divides k 1 k + 1 and not k 3 k and so 3 holds and 14 follows. Finally, we shall prove 17. Let c 3,c 4,... denote positive numbers which are effectively computable in terms of ε and δ. Without loss of generality we may suppose that n i 1 > i > > i t 1.
8 8 C.L. Stewart Note that we may also suppose that since otherwise, by 16, and therefore, by 14, t 1 log n 1 8, 7 llog n t 1 llog 1 n3 4 nlog n 4 A < exp llog n 1 + ε t 1, for n larger than c 3, whence 17 holds. We consider the consecutive integers i j+1 +1,...,i j for j = 1,...,t 1. Notice that A divides i j!+1 and i j+1!+1 hence A divides i j i j Therefore A is at most n i j i j+1. If for some j, with 1 j t 1, n i j i j+1 <, tlog n 3 then n A exp, tlog n 1 and, by 16, 17 holds. Thus we may suppose that n i j i j+1, tlog n 3 for j = 1,...,t 1. Let m denote the number of the intervals [i j+1 +1,i j ] for j = 1,...,t 1 which do not contain a prime number. Let p 1,p,... denote the sequence of prime numbers and put d k = p k+1 p k for k = 1,,.... Then d k m n. tlog n 3 But, by Lemma 1, for n > c 4. In particular p k n p k n d k < n δ, m < t log n 3, n δ
9 and by 15, since t n, On the greatest and least prime factors of n! + 1, II 9 m < tn δ log n 3 < t 1 δ 3, 8 for n > c 5. Put t 1 = t 1 m, 9 and order the differences i j i j+1 with 1 j t 1 for which there is a prime in the interval [i j+1 + 1,i j ] according to size. Let us denote these differences by γ 1,...,γ t1 so that γ 1 γ γ t1. 30 Observe that γ γ t1 i 1 i t l. 31 For any real number x let [x] denote the largest integer of size at most x. Put t = [t 1 /log log t 1 1 ]. 3 Then, by 30 and 31, γ t t 1 t l. Thus, by 7, 8, 9, 30 and 3, for n > c 6, for h = 1,...,t. Next note that γ h < 1 + ε l t, 33 γ t γ t 1 + γ t 1 γ t + +γ 1 0 = γ t γ t 1 γ t + +γ 1 0 = γ t hence γ 1 0 = γ 1 γ t γ t 1 + γ t 1 γ t + + t γ 1 = γ t + + γ 1. 34
10 10 C.L. Stewart Put θ = minγ t γ t 1,...,γ γ 1,γ 1. Then, by 31 and 34, t t + 1 θ l. Therefore, by 7, 8, 9 and 3, for n > c 7, llog log t θ < 1 + ε t. 35 We have γ 1 = i r i r+1 for an integer r with 1 r t 1. Then A i r i r If θ = γ 1, we see that A < n θ, and by 35 our result follows. Thus we may suppose that θ = γ s γ s 1 for some integer s from {,...,t }. In particular, θ = i a i a+1 i b i b+1 with a and b distinct integers from {1,...,t 1}. Put k 1 = i a, k = i a+1, k 3 = i b and k 4 = i b+1. By construction there is a prime among the integers k + 1,...,k 1 and another prime among the integers k 4 + 1,...,k 3. Thus the larger of the two primes divides one of k 1 k +1 and k 3 k and not the other whence 3 holds. Note also that ne x x is an increasing function of x for x positive and less than n. Therefore, by 4, 7, 33 and 35, we find that log log t 1+εl net 1+ε l t A < n t, 1 + εl hence that A < exp 1 + εl log n log log t t + log t t log ne 1+εl + 36 t for n > c 8, as required.
11 On the greatest and least prime factors of n! + 1, II 11 For any prime p let tp denote the number of positive integers k for which p k! + 1. In [9, Theorem 7.5] we noted that tp < m + 1m + where m = 3p To see this observe that if n and s are positive integers and p divides both n!+1 and n+s!+1 then p divides n+s! n! hence n+s n+1 1 mod p. In particular, n is a solution of the polynomial congruence x + s x mod p, and by Lagrange s theorem the number of such solutions is at most s. Let I be an interval of length l 1 and let n 1 < n < < n k denote all the solutions of x! mod p in I. Plainly k 1 n i+1 n i l, 38 i=1 and by our earlier observation at most s of the terms in brackets in the above sum are equal to s. Therefore k 1 n i+1 n i i=1 where u is defined by the inequalities u s, 39 s=1 u u+1 s k 1 < s. s=1 s=1 Thus k is at most u+1u+ and by 38 and 39 l uu + 1u > u Since all integers n for which p n! + 1 lie in the interval [1,p 1], 37 follows from 40 with l = p. Further, from 40 we obtain Lemma 3 below, a result which is the special case of Lemma of Luca and Shparlinski [7] with fx equal to 1. Lemma 3. There exists a positive number c such that if p is a prime number and I is an interval of the positive real numbers of length l with l 1 then the number of integers k in I for which p divides k! + 1 is at most cl 3.
12 1 C.L. Stewart For any nonzero integer m and any prime p we denote by ord p m the exponent of the largest power of p which divides m. As usual m p is the padic absolute value of m normalized so that m p = p ordpm. Lemma 4. There exists a positive number c 1 such that if p is a prime number, n a positive integer and I a subinterval of [1,n] of length l then log pord p i Ii! + 1 < 3 llog llog n + c 1llog n + n log n. 41 Further, for each pair of positive real numbers ε and ε 1 there exist positive numbers c and c 3 such that if l exceeds ε 1 n and n exceeds c 3 then log pord p + 1 < 1 + ε i Ii! 9 llog l + c n log n. 4 Proof. Let i 1,...,i u be the integers i in I for which p divides i! + 1. By Lemma 3 there is a positive number c such that u cl Put h t = ord p i t! + 1 for t = 1,...,u. We may suppose that Then, by 13 of Lemma, h 1 h u. p ht < n l t 1, 44 for t =,...,u. In particular by 43 and 44, log ph + + h u < llog n 1 + cl 3 1 t 1 dt < 3 llog llog n + c 4llog n, 45
13 On the greatest and least prime factors of n! + 1, II 13 where c 4 is a positive number. Since h 1 log p n log n 46 and ord p i Ii! + 1 = h h u, follows from 45 and 46. Let c 5,c 6,... denote positive numbers which depend on ε and ε 1 and suppose that l exceeds ε 1 n. Then by 43, u cn 3 < n , for n > c 5. Thus for n > c 6, 15 of Lemma holds with δ = 1 36 and, since l > ε 1 n, 16 of Lemma also holds. Therefore by 17 of Lemma, for n > c 7, log ph t < 1 + εl log t log e ε 1 log n max1,log log t + + t t t 1 48 for t = 3,...,u. Since the expression on the right hand side of 48 is a decreasing function of t for t > e, we see that and so log ph h u cl3 < 1 + εl 3 log t t + log e ε 1 t + log n max1,log log t t 1 dt log ph h u < 1 + ε 9 llog l + c 8 n log n. 49 Since h 1 + h + h 3 log p < 3n log n, 50 4 now follows from 47, 49 and 50. Lemma 5. Let ε be a positive real number. There exists a positive number c, which depends on ε, such that if p is a prime number, n an integer with n and I a subinterval of [1,n] of length l then log p ord p i Ii! + 1 < 9 llog n + εnlog n + cn log n. 51
14 14 C.L. Stewart Proof. Let c 1,c,... denote positive numbers which depend on ε. By 4 of Lemma 4, if l exceeds εn and n exceeds c 1, log p ord p + 1 < i Ii! 9 llog n + 9 εnlog n + c n log n. 5 On the other hand if l εn then by 41 of Lemma 4, log p ord p + 1 < i Ii! 3 εnlog n + c 3 n log n; 53 plainly 53 holds if l. Therefore from 5 and 53, we obtain 51 with c 4 in place of c for n > c 1, hence 51 holds for n and our result follows. 3. Proof of Theorem 1 Let δ be a positive real number with δ < Put δ = 10δ, λ = + δ, 54 8 and λ 1 = λ + δ. Note that λ < 3. Let c 1,c,... denote positive numbers which depend on δ. We may suppose that there exist only finitely many odd positive integers n for which pn! + 1 λ 1 n, since otherwise the theorem holds. Thus there exists a positive integer c 1 such that for each odd integer n with n > c 1, pn! + 1 > λ 1 n. 55 We shall show that this leads to a contradiction and the theorem then follows. Since 55 holds, we also have Pn! + 1 < λ n, 56 λ 1
15 On the greatest and least prime factors of n! + 1, II 15 for all odd integers n with n > c. To see this, observe that if q = Pn!+1 with n > 1 and q λ n, 57 λ 1 then q is odd and, by 4, q q n 1! + 1. But then 1 pq n 1! + 1 q = 1 n+1 q n 1. q We have q > λ and, by 57, n q λ 1 λ hence 1 1 n+1 q q = qλ λ 1 q λ. λ But qλ q λ < λ 1 for n > c 3 since q > n. Thus pq n 1! + 1 λ 1 q n 1. Furthermore, q n 1 > c 1 for n > c 4 by 57 and this contradicts 55. Therefore 56 holds. The proof now proceeds by a comparison of estimates for Z = N n! + 1. n=1 n odd, n>c Put R = {n Z n odd, c < n N}. Observe that if p n! + 1 with n in R then, by 55, p > λ 1 n and, by 56, p < λ λ 1n. In particular, Put Since n! n e n, I p = λ 1 λ p < n < 1 λ 1 p. λ 1 λ p,min N, 1 p. λ 1 Z > exp n log n n n R
16 16 C.L. Stewart so log Z > 1 δ N log N, 58 4 provided that N exceeds c 5. On the other hand Z = Z 1 p 1 n! + 1. p p Put n I p R Ap = log pord p n I p R p n! + 1. Then Z exp Ap. p< λ λ 1 N Thus by 51 of Lemma 5, with ε = δ, log Z log N 9 p< λ λ 1 N lp + δnlog N + c 6 N log Nπ λ λ 1 N, 59 where lp, the length of I p, is given by lp = 1 λ 1 p when p λ 1 N λ 1 λ and by λ 1 lp = N p when p λ 1 N. λ By 54, 59 and the prime number theorem, log Z log N 9 p< λ λ 1 N lp + 4δN log N + c 7 N. 60
17 Further p< λ λ 1 N lp = On the greatest and least prime factors of n! + 1, II 17 p λ 1 N = 1 λ 1 1 λ 1 p + λ 1 λ p λ 1 N p + N λ 1 N<p< λ λ 1 N λ 1 N<p< λ λ 1 N 1 N λ 1 λ p λ 1 λ p< λ λ 1 N p. Thus by the prime number theorem and Abel summation, for N > c 8, λ1 λ lp < 1 + δ + λ 1 λ λ N 1 λ 1 log N p< λ λ 1 N λ < 1 + δ λ 1 λ 1 N log N, and so by 60, and the fact that λ 1 exceeds λ, 1 + δ λ log Z < 9 λ 1 λ N log N + 4δN log N + c 7 N. 61 Comparing 58 and 61 we find that for N > c 9, 1 δ 4 < 1 + δ 9 λ λ λ 1 + 4δ + c 8 log N. By 54 we obtain a contradiction for N sufficiently large and the result now follows. 4. Proof of Theorem We may suppose that 0 < ε < Put γ = the set of positive integers for which 18ε and let Bγ be Pn! + 1 γn. 6 We shall show that for n sufficiently large the set Bγ {1,...,n} has cardinality at least ε 3n and hence the result follows. Accordingly suppose that N is a positive integer for which Bγ {1,...,N} ε N. 63 3
18 18 C.L. Stewart Let c 1,c,... denote positive numbers which depend on ε. Our proof proceeds by a comparison of estimates for Z = N n=1 n Bγ n! + 1. Since n! n n e, N Z > exp n log n n Bγ {1,...,N} N log N whence, by 63, n= log Z > 1 ε N log N, 64 provided that N > c 1. Notice that if p n! + 1 and n Bγ then n < p < γn hence 1 γ p < n < p. Put I p = and Then 1 γ p,min{n,p}, Ap = log pord p n Ipn! + 1. Z exp Ap. p<γn Thus, by 51 of Lemma 5 with ε 6 in place of ε, log Z < ε log N lp Nlog N + c N log N πγn, 65 p<γn where lp, the length of I p, is given by lp = 1 1 p for p N γ and lp = N 1 p for p > N. γ
19 Since γ < 11 On the greatest and least prime factors of n! + 1, II 19 it follows from 65 and the prime number theorem that log Z < log N lp + εn log N + c 3 N p<γn Further p<γn lp = p N = p N 1 1 p + γ p + N<p<γN N<p<γN N 1 γ N 1γ p p<γn p. Thus, by the prime number theorem and Abel summation, for N > c 4, N γ 1N lp < 1 + ε + log N log N γ N log N p<γn and so by 66, log Z < 1 + ε γ 1N log N + εn log N + c 3 N Comparing 64 and 67 we find that for N > c 5, 1 ε γ 1 < 1 + ε + ε + c 3 9 log N. But γ < 11 and so for N sufficiently large we obtain a contradiction. Thus 63 does not hold for N sufficiently large and the result now follows. References [1] B. Brindza and P. Erdős, On some Diophantine problems involving powers and factorials, J. Austral. Math. Soc. Ser. A , 1 7. [] P. Erdős and R. Graham, Old and new problems and results in combinatorial number theory, Monographie 8 de L Enseignement Mathématique, Genève, [3] P. Erdős and C.L. Stewart, On the greatest and least prime factors of n! + 1, J. London Math. Soc 13, no. 1976, [4] D.R. HeathBrown, The difference between consecutive primes, III, J. London Math. Soc. 0, no. 1979,
20 0 C.L. Stewart [5] M. Le, On the Diophantine equation x p 1 + p 1! = p n, Publ. Math. Debrecen , [6] J. Liouville, Sur l équation 1 3 p = p m, J. Math. Pure Appl , [7] F. Luca and I.E. Shparlinski, Prime divisors of shifted factorials to appear. [8] M.R. Murty and S. Wong, The ABC conjecture and prime divisors of the Lucas and Lehmer sequences, Number Theory for the Millenium, III, Urbana, IL, 000, A.K. Peters, Natick, MA, 00, [9] C.L. Stewart, On divisor properties of arithmetical sequences, Ph.D. thesis, University of Cambridge, [10] K. Yu and D. Liu, A complete resolution of a problem of Erdös and Graham, Rocky Mountain J. Math , DEPARTMENT OF PURE MATHEMATICS UNIVERSITY OF WATERLOO WATERLOO, ONTARIO CANADA NL 3G1 URL:
PRIME FACTORS OF CONSECUTIVE INTEGERS
PRIME FACTORS OF CONSECUTIVE INTEGERS MARK BAUER AND MICHAEL A. BENNETT Abstract. This note contains a new algorithm for computing a function f(k) introduced by Erdős to measure the minimal gap size in
More informationOn the largest prime factor of the Mersenne numbers
On the largest prime factor of the Mersenne numbers Kevin Ford Department of Mathematics The University of Illinois at UrbanaChampaign Urbana Champaign, IL 61801, USA ford@math.uiuc.edu Florian Luca Instituto
More informationFACTORING SPARSE POLYNOMIALS
FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix nonzero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S
More informationOn the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y = (5m) z
International Journal of Algebra, Vol. 9, 2015, no. 6, 261272 HIKARI Ltd, www.mhikari.com http://dx.doi.org/10.12988/ija.2015.5529 On the Exponential Diophantine Equation (12m 2 + 1) x + (13m 2 1) y
More informationl4 f0 (n) 2 f,(n) > f2 (n) 2 f 5 (n) 2 f4 (n) z f5 (n). CONJECTURE : It seems certain to us that for infinitely many n the inequalities are all strict
SOME PROBLEMS ON 'THE PRIME FACTORS OF CONSECUTIVE, INTEGERS II by P. Erdös and J. L. Selfridge G. A. Grimm [3) stated the following interesting conjecture : Let n + 1,..., n + k be consecutive composite
More informationON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM
Acta Math. Univ. Comenianae Vol. LXXXI, (01), pp. 03 09 03 ON INTEGERS EXPRESSIBLE BY SOME SPECIAL LINEAR FORM A. DUBICKAS and A. NOVIKAS Abstract. Let E(4) be the set of positive integers expressible
More informationNotes on the second moment method, Erdős multiplication tables
Notes on the second moment method, Erdős multiplication tables January 25, 20 Erdős multiplication table theorem Suppose we form the N N multiplication table, containing all the N 2 products ab, where
More information5 The Beginning of Transcendental Numbers
5 The Beginning of Transcendental Numbers We have defined a transcendental number (see Definition 3 of the Introduction), but so far we have only established that certain numbers are irrational. We now
More informationPrimitive Prime Divisors of FirstOrder Polynomial Recurrence Sequences
Primitive Prime Divisors of FirstOrder Polynomial Recurrence Sequences Brian Rice brice@hmc.edu July 19, 2006 Abstract The question of which terms of a recurrence sequence fail to have primitive prime
More informationThe minimal number with a given number of. divisors
The minimal number with a given number of divisors Ron Brown Department of Mathematics, 2565 McCarthy Mall, University of Hawaii, Honolulu, HI 96822, USA Phone (808) 9564665 Fax (808) 9569139 Email address:
More informationOn the Union of Arithmetic Progressions
On the Union of Arithmetic Progressions Shoni Gilboa Rom Pinchasi August, 04 Abstract We show that for any integer n and real ɛ > 0, the union of n arithmetic progressions with pairwise distinct differences,
More informationOn the numbertheoretic functions ν(n) and Ω(n)
ACTA ARITHMETICA LXXVIII.1 (1996) On the numbertheoretic functions ν(n) and Ω(n) by Jiahai Kan (Nanjing) 1. Introduction. Let d(n) denote the divisor function, ν(n) the number of distinct prime factors,
More informationAPPLICATIONS 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
More informationConcentration of points on two and three dimensional modular hyperbolas and applications
Concentration of points on two and three dimensional modular hyperbolas and applications J. Cilleruelo Instituto de Ciencias Matemáticas (CSICUAMUC3MUCM) and Departamento de Matemáticas Universidad
More informationOn the largest prime factor of x 2 1
On the largest prime factor of x 2 1 Florian Luca and Filip Najman Abstract In this paper, we find all integers x such that x 2 1 has only prime factors smaller than 100. This gives some interesting numerical
More informationChapter 1 Basic Number Concepts
Draft of September 2014 Chapter 1 Basic Number Concepts 1.1. Introduction No problems in this section. 1.2. Factors and Multiples 1. Determine whether the following numbers are divisible by 3, 9, and 11:
More informationOn fractional parts of powers of real numbers close to 1
On fractional parts of powers of real numbers close to 1 Yann BUGEAUD and Niolay MOSHCHEVITIN Abstract We prove that there exist arbitrarily small positive real numbers ε such that every integral power
More informationCHAPTER 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.
More informationIntroduction to Series and Sequences Math 121 Calculus II D Joyce, Spring 2013
Introduction to Series and Sequences Math Calculus II D Joyce, Spring 03 The goal. The main purpose of our study of series and sequences is to understand power series. A power series is like a polynomial
More informationCharacterizing the Sum of Two Cubes
1 3 47 6 3 11 Journal of Integer Sequences, Vol. 6 (003), Article 03.4.6 Characterizing the Sum of Two Cubes Kevin A. Broughan University of Waikato Hamilton 001 New Zealand kab@waikato.ac.nz Abstract
More information1. 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
More informationPrime 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
More informationPrime Numbers and Irreducible Polynomials
Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.
More informationU.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
More informationHomework 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
More informationConvergence and stability results for a class of asymptotically quasinonexpansive mappings in the intermediate sense
Functional Analysis, Approximation and Computation 7 1) 2015), 57 66 Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/faac Convergence and
More informationPERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES. 1. Introduction
PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES LINDSEY WITCOSKY ADVISOR: DR. RUSS GORDON Abstract. This paper examines two methods of determining whether a positive integer can correspond to the semiperimeter
More informationEvery Positive Integer is the Sum of Four Squares! (and other exciting problems)
Every Positive Integer is the Sum of Four Squares! (and other exciting problems) Sophex University of Texas at Austin October 18th, 00 Matilde N. Lalín 1. Lagrange s Theorem Theorem 1 Every positive integer
More informationThe squarefree kernel of x 2n a 2n
ACTA ARITHMETICA 101.2 (2002) The squarefree kernel of x 2n a 2n by Paulo Ribenboim (Kingston, Ont.) Dedicated to my longtime friend and collaborator Wayne McDaniel, at the occasion of his retirement
More informationTHE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. Saturday, March 5, 2016
THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION In Memory of Robert Barrington Leigh Saturday, March 5, 2016 Time: 3 1 2 hours No aids or calculators permitted. The grading is designed
More informationFACTORS OF CARMICHAEL NUMBERS AND A WEAK ktuples CONJECTURE. 1. Introduction Recall that a Carmichael number is a composite number n for which
FACTORS OF CARMICHAEL NUMBERS AND A WEAK ktuples CONJECTURE THOMAS WRIGHT Abstract. In light of the recent work by Maynard and Tao on the Dickson ktuples conjecture, we show that with a small improvement
More informationPrime divisors of Lucas sequences and a conjecture of Ska lba
Prime divisors of Lucas sequences and a conjecture of Ska lba Florian Luca, Instituto de Matemáticas Universidad Nacional Autónoma de México C.P. 58089, Morelia, Michoacán, México fluca@matmor.unam.mx
More informationCONTRIBUTIONS TO ZERO SUM PROBLEMS
CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg
More informationCOMPOSITES THAT REMAIN COMPOSITE AFTER CHANGING A DIGIT
COMPOSITES THAT REMAIN COMPOSITE AFTER CHANGING A DIGIT Michael Filaseta Mathematics Department University of South Carolina Columbia, SC 22080001 Mark Kozek Department of Mathematics Whittier College
More informationCHAPTER 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
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More informationON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS
ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK Abstract If n is a positive integer, write F n for the nth Fibonacci number, and ω(n) for the number
More informationHOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!
Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following
More informationThe sum of digits of polynomial values in arithmetic progressions
The sum of digits of polynomial values in arithmetic progressions Thomas Stoll Institut de Mathématiques de Luminy, Université de la Méditerranée, 13288 Marseille Cedex 9, France Email: stoll@iml.univmrs.fr
More informationIntroduction to Finite Fields (cont.)
Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number
More informationCONTINUED 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.
More informationLecture 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
More informationarxiv:math/ v2 [math.nt] 6 Aug 2006
PROBLEMS IN ADDITIVE NUMBER THEORY, I arxiv:math/0604340v2 [math.nt] 6 Aug 2006 MELVYN B. NATHANSON Abstract. Talk at the Atelier en combinatoire additive (Workshop on Arithmetic Combinatorics) at the
More informationk, 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
More informationErdos and the Twin Prime Conjecture: Elementary Approaches to Characterizing the Differences of Primes
Erdos and the Twin Prime Conjecture: Elementary Approaches to Characterizing the Differences of Primes Jerry Li June, 010 Contents 1 Introduction Proof of Formula 1 3.1 Proof..................................
More informationMath 55: Discrete Mathematics
Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What
More informationMOP 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
More informationN E W S A N D L E T T E R S
N E W S A N D L E T T E R S 73rd Annual William Lowell Putnam Mathematical Competition Editor s Note: Additional solutions will be printed in the Monthly later in the year. PROBLEMS A1. Let d 1, d,...,
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More informationIB Math 11 Assignment: Chapters 1 & 2 (A) NAME: (Functions, Sequences and Series)
IB Math 11 Assignment: Chapters 1 & 2 (A) NAME: (Functions, Sequences and Series) 1. Let f(x) = 7 2x and g(x) = x + 3. Find (g f)(x). Write down g 1 (x). (c) Find (f g 1 )(5). (Total 5 marks) 2. Consider
More informationSequences of Functions
Sequences of Functions Uniform convergence 9. Assume that f n f uniformly on S and that each f n is bounded on S. Prove that {f n } is uniformly bounded on S. Proof: Since f n f uniformly on S, then given
More informationLOWDEGREE PLANAR MONOMIALS IN CHARACTERISTIC TWO
LOWDEGREE PLANAR MONOMIALS IN CHARACTERISTIC TWO PETER MÜLLER AND MICHAEL E. ZIEVE Abstract. Planar functions over finite fields give rise to finite projective planes and other combinatorial objects.
More informationCycles in a Graph Whose Lengths Differ by One or Two
Cycles in a Graph Whose Lengths Differ by One or Two J. A. Bondy 1 and A. Vince 2 1 LABORATOIRE DE MATHÉMATIQUES DISCRÉTES UNIVERSITÉ CLAUDEBERNARD LYON 1 69622 VILLEURBANNE, FRANCE 2 DEPARTMENT OF MATHEMATICS
More informationFIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
More informationPrime and Composite Terms in Sloane s Sequence A056542
1 2 3 47 6 23 11 Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.3 Prime and Composite Terms in Sloane s Sequence A056542 Tom Müller Institute for CusanusResearch University and Theological
More information4.3 Limit of a Sequence: Theorems
4.3. LIMIT OF A SEQUENCE: THEOREMS 5 4.3 Limit of a Sequence: Theorems These theorems fall in two categories. The first category deals with ways to combine sequences. Like numbers, sequences can be added,
More informationk=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =
Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem
More informationOn Amicable and Sociable Numbers *
MATHEMATICS OF COMPUTATION. VOLUME 24. NUMBER 110, APRIL, 1970 On Amicable and Sociable Numbers * By Henri Cohen Abstract. An exhaustive search has yielded 236 amicable pairs of which the lesser number
More informationarxiv: v1 [math.pr] 22 Aug 2008
arxiv:0808.3155v1 [math.pr] 22 Aug 2008 On independent sets in purely atomic probability spaces with geometric distribution Eugen J. Ionascu and Alin A. Stancu Department of Mathematics, Columbus State
More informationCourse 221: Analysis Academic year , First Semester
Course 221: Analysis Academic year 200708, First Semester David R. Wilkins Copyright c David R. Wilkins 1989 2007 Contents 1 Basic Theorems of Real Analysis 1 1.1 The Least Upper Bound Principle................
More informationReal quadratic fields with class number divisible by 5 or 7
Real quadratic fields with class number divisible by 5 or 7 Dongho Byeon Department of Mathematics, Seoul National University, Seoul 151747, Korea Email: dhbyeon@math.snu.ac.kr Abstract. We shall show
More informationDiscrete Mathematics, Chapter 5: Induction and Recursion
Discrete Mathematics, Chapter 5: Induction and Recursion Richard Mayr University of Edinburgh, UK Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. Chapter 5 1 / 20 Outline 1 Wellfounded
More informationCounting Primes whose Sum of Digits is Prime
2 3 47 6 23 Journal of Integer Sequences, Vol. 5 (202), Article 2.2.2 Counting Primes whose Sum of Digits is Prime Glyn Harman Department of Mathematics Royal Holloway, University of London Egham Surrey
More informationSimultaneous equal sums of three powers
Simultaneous equal sums of three powers T.D. Browning 1 and D.R. HeathBrown 2 1 School of Mathematics, Bristol University, Bristol BS8 1TW 2 Mathematical Institute, 24 29 St. Giles, Oxford OX1 3LB 1 t.d.browning@bristol.ac.uk,
More informationRearranging the Alternating Harmonic Series
Rearranging the Alternating Harmonic Series Lawrence H. Riddle Department of Mathematics Agnes Scott College, Decatur, GA 30030 Why are conditionally convergent series interesting? While mathematicians
More informationSolutions 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
More informationTwogenerator numerical semigroups and Fermat and Mersenne numbers
Twogenerator numerical semigroups and Fermat and Mersenne numbers Shalom Eliahou and Jorge Ramírez Alfonsín Abstract Given g N, what is the number of numerical semigroups S = a,b in N of genus N\S = g?
More informationThe fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992), 33 37) Bart de Smit
The fundamental group of the Hawaiian earring is not free Bart de Smit The fundamental group of the Hawaiian earring is not free (International Journal of Algebra and Computation Vol. 2, No. 1 (1992),
More informationModule MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions
Module MA1S11 (Calculus) Michaelmas Term 2016 Section 3: Functions D. R. Wilkins Copyright c David R. Wilkins 2016 Contents 3 Functions 43 3.1 Functions between Sets...................... 43 3.2 Injective
More informationChapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1
Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes
More informationPrimes from the viewpoint of combinatorics
A talk given at Jiangsu Normal University (March 22, 2012) Workshop in Combinatorics and Graph (Changsha, June 10, 2012) The 5th National Conferfence on Combinatorics and Graph theory (Luoyang, July 18,
More informationGREATEST COMMON DIVISOR
DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their
More informationSOME TYPES OF CONVERGENCE OF SEQUENCES OF REAL VALUED FUNCTIONS
Real Analysis Exchange Vol. 28(2), 2002/2003, pp. 43 59 Ruchi Das, Department of Mathematics, Faculty of Science, The M.S. University Of Baroda, Vadodara  390002, India. email: ruchid99@yahoo.com Nikolaos
More informationON THE COEFFICIENTS OF THE CYCLOTOMIC POLYNOMIAL
ON THE COEFFICIENTS OF THE CYCLOTOMIC POLYNOMIAL PAUL ERDÖS The cyclotomic polynomial F n (x) is defined as the polynomial whose roots are the primitive nth roots of unity. It is well known that Fn(x)
More informationON TOPOLOGICAL SEQUENCE ENTROPY AND CHAOTIC MAPS ON INVERSE LIMIT SPACES. 1. Introduction
Acta Math. Univ. Comenianae Vol. LXVIII, 2(1999), pp. 205 211 205 ON TOPOLOGICAL SEQUENCE ENTROPY AND CHAOTIC MAPS ON INVERSE LIMIT SPACES J. S. CANOVAS Abstract. The aim of this paper is to prove the
More informationp 2 1 (mod 6) Adding 2 to both sides gives p (mod 6)
.9. Problems P10 Try small prime numbers first. p p + 6 3 11 5 7 7 51 11 13 Among the primes in this table, only the prime 3 has the property that (p + ) is also a prime. We try to prove that no other
More informationCHAPTER III  MARKOV CHAINS
CHAPTER III  MARKOV CHAINS JOSEPH G. CONLON 1. General Theory of Markov Chains We have already discussed the standard random walk on the integers Z. A Markov Chain can be viewed as a generalization of
More informationMA257: 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
More informationON THE FIBONACCI NUMBERS
ON THE FIBONACCI NUMBERS Prepared by Kei Nakamura The Fibonacci numbers are terms of the sequence defined in a quite simple recursive fashion. However, despite its simplicity, they have some curious properties
More informationMathematical 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
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationCourse Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction.
Course Notes for Math 320: Fundamentals of Mathematics Chapter 3: Induction. February 21, 2006 1 Proof by Induction Definition 1.1. A subset S of the natural numbers is said to be inductive if n S we have
More informationTopics in Number Theory
Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall
More informationMATH 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,
More informationFibonacci Numbers Modulo p
Fibonacci Numbers Modulo p Brian Lawrence April 14, 2014 Abstract The Fibonacci numbers are ubiquitous in nature and a basic object of study in number theory. A fundamental question about the Fibonacci
More informationPrimality  Factorization
Primality  Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.
More informationTakeAway Games MICHAEL ZIEVE
Games of No Chance MSRI Publications Volume 29, 1996 TakeAway Games MICHAEL ZIEVE Abstract. Several authors have considered takeaway games where the players alternately remove a positive number of counters
More information0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2.
SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then xy x + y. Proof. First we prove that if x is a real number, then x 0. The product of two positive
More informationSection 5.1 Number Theory: Prime & Composite Numbers
Section 5.1 Number Theory: Prime & Composite Numbers Objectives 1. Determine divisibility. 2. Write the prime factorization of a composite number. 3. Find the greatest common divisor of two numbers. 4.
More informationSeries. Chapter Convergence of series
Chapter 4 Series Divergent series are the devil, and it is a shame to base on them any demonstration whatsoever. Niels Henrik Abel, 826 This series is divergent, therefore we may be able to do something
More informationAll trees contain a large induced subgraph having all degrees 1 (mod k)
All trees contain a large induced subgraph having all degrees 1 (mod k) David M. Berman, A.J. Radcliffe, A.D. Scott, Hong Wang, and Larry Wargo *Department of Mathematics University of New Orleans New
More informationMathematical Induction. Lecture 1011
Mathematical Induction Lecture 1011 Menu Mathematical Induction Strong Induction Recursive Definitions Structural Induction Climbing an Infinite Ladder Suppose we have an infinite ladder: 1. We can reach
More informationMath 1B, lecture 14: Taylor s Theorem
Math B, lecture 4: Taylor s Theorem Nathan Pflueger 7 October 20 Introduction Taylor polynomials give a convenient way to describe the local behavior of a function, by encapsulating its first several derivatives
More informationIrreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients
DOI: 10.2478/auom20140007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca
More informationMore Mathematical Induction. October 27, 2016
More Mathematical Induction October 7, 016 In these slides... Review of ordinary induction. Remark about exponential and polynomial growth. Example a second proof that P(A) = A. Strong induction. Least
More informationCHAPTER IV NORMED LINEAR SPACES AND BANACH SPACES
CHAPTER IV NORMED LINEAR SPACES AND BANACH SPACES DEFINITION A Banach space is a real normed linear space that is a complete metric space in the metric defined by its norm. A complex Banach space is a
More informationCharacterization of arithmetic functions that preserve the sumofsquares operation
Characterization of arithmetic functions that preserve the sumofsquares operation Bojan Bašić Department of Mathematics and Informatics, University of Novi Sad, Trg Dositeja Obradovića, 1000 Novi Sad,
More informationA DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS
A DIOPHANTINE EQUATION RELATED TO THE SUM OF POWERS OF TWO CONSECUTIVE GENERALIZED FIBONACCI NUMBERS ANA PAULA CHAVES AND DIEGO MARQUES Abstract. Let F n) n 0 be the Fibonacci sequence given by F m+ =
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient
More information(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map
22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find
More information