# MODULAR ARITHMETIC (SHORT VERSION)

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1 MODULAR ARITHMETIC (SHORT VERSION) KEITH CONRAD. Introduction We will define the notion of congruent integers (with respect to a modulus) and develop some basic ideas of modular arithmetic, which lets us carry out algebraic calculations on integers with a systematic disregard for terms divisible by a certain number (called the modulus). This form of reduced algebra is essential background for the mathematics of computer science, coding theory, cryptography, primality testing, and much more. 2. Integer congruences The following definition was introduced by Gauss in his Disquisitiones Arithmeticae (Arithmetic Investigations) in 80. Definition 2. (Gauss). Let m be an integer. For a, b Z, we write a b mod m and say a is congruent to b modulo m if m (a b). Example 2.2. Check 8 4 mod 7, 20 3 mod, and 9 3 mod 2. Remark 2.3. The parameter m is called the modulus, not the modulo. We have m 0 mod m, and more generally mk 0 mod m for any k Z. In fact, a 0 mod m m a, so the congruence relation includes the divisibility relation as a special case: multiples of m are exactly the numbers that look like 0 modulo m. Because multiples of m are congruent to 0 modulo m, we will see that working with integers modulo m amounts to systematically ignoring additions and subtractions by multiples of m in calculations. Since a b mod m if and only if b = a + mk for some k Z, adjusting an integer modulo m is the same as adding (or subtracting) multiples of m to it. Thus, if we want to find a positive integer congruent to 8 mod 5, we can add a multiple of 5 to 8 until we go positive. Adding 20 does the trick: = 2, so 8 2 mod 5 (check!). There is a useful analogy between integers modulo m and angle measurements (in radians, say). In both cases, the objects involved admit different representations, e.g., the angles 0, 2π, and 4π are the same, just as mod 5. Every angle can be put in standard form as a real number in the interval [0, 2π). There is a similar convention for the standard representation of an integer modulo m using remainders, as follows. Theorem 2.4. Let m Z be a nonzero integer. For each a Z, there is a unique r with a r mod m and 0 r < m.

2 2 KEITH CONRAD Proof. Using division with remainder in Z, there are q and r in Z such that a = mq + r, 0 r < m. Then m (a r), so a r mod m. To show r is the unique number in the range {0,,..., m } that is congruent to a mod m, suppose two numbers in this range work: where 0 r, r < m. Then we have a r mod m, a r mod m, a = r + mk, a = r + ml for some k and l in Z, so the remainders r and r have difference r r = m(l k). This is a multiple of m, and the bounds on r and r tell us r r < m. A multiple of m has absolute value less than m only if it is 0, so r r = 0, which means r = r. Example 2.5. Taking m = 2, every integer is congruent modulo 2 to exactly one of 0 and. Saying n 0 mod 2 means n = 2k for some integer k, so n is even, and saying n mod 2 means n = 2k + for some integer k, so n is odd. We have a b mod 2 precisely when a and b have the same parity: both are even or both are odd. Example 2.6. Every integer is congruent mod 4 to exactly one of 0,, 2, or 3. Congruence mod 4 is a refinement of congruence mod 2: even numbers are congruent to 0 or 2 mod 4 and odd numbers are congruent to or 3 mod 4. For instance, 0 2 mod 4 and 9 3 mod 4. Congruence mod 4 is related to Master Locks. Every combination on a Master Lock is a triple of numbers (a, b, c) where a, b, and c vary from 0 to 39. Each number has 40 choices, with b a and c b (perhaps c = a). This means the number of combinations could be up to 60840, but in fact the true number of combinations is a lot smaller: every combination has c a mod 4 and b a + 2 mod 4, so once some number in a combination is known the other two numbers are each limited to 0 choices (among the 40 numbers in {0,,..., 39} exactly 0 will be congruent to a particular value mod 4). Thus the real number of Master Lock combinations is = Example 2.7. Taking m = 7, every integer is congruent modulo 7 to exactly one of 0,, 2,..., 6. The choice is the remainder when the integer is divided by 7. For instance, 20 6 mod 7 and 32 3 mod 7. Definition 2.8. We call {0,, 2,..., m } the standard representatives for integers modulo m. In practice m > 0, so the standard representatives modulo m are {0,, 2,..., m }. In fact, congruence modulo m and modulo m are the same relation (just look back at the definition), so usually we never talk about negative moduli. Nevertheless, Theorem 2.4 is stated for any nonzero modulus m. By Theorem 2.4, there are m incongruent integers modulo m. Each integer is congruent modulo m to a standard representative, just like any fraction can be written in a reduced form. There are many other representatives, however, and this will be important!

4 4 KEITH CONRAD Other notations you may meet for Z/(m) are Z m and Z/mZ. Example 3.7. Here are the elements of Z/(5): {..., 5, 0, 5, 0, 5, 0, 5,... }, {..., 4, 9, 4,, 6,, 6,... }, {..., 3, 8, 3, 2, 7, 2, 7,... }, {..., 2, 7, 2, 3, 8, 3, 8,... }, {...,, 6,, 4, 9, 4, 9,... }. These five sets each consist of all the integers congruent to each other modulo 5, so each set is called a congruence class (modulo 5). In practice we often use one representative from each congruence class to stand for the whole congruence class. In bold type is one set of representatives for the congruence classes modulo 5 the choice 0,, 2, 3, and 4 which is the standard one. Another set of representatives is 5, 6, 3, 8, and 6 (since 5 0 mod 5, 6 mod 5, 3 2 mod 5, and so on). Different integers in the same congruence class are like different real numbers representing the same angle, such as π and 3π. They re different ways of representing the same thing. Just as π and 3π are not the same as real numbers but become the same in the setting of angles, we need to think this way about congruence classes: and 6 are not the same in Z but they are the same when we think modulo Solving equations in Z/(m) In school you solve equations like 2x + 3 = 8 or x 2 3x + = 0 by the rules of algebra : cancellation (if a 0 and ax = ay then x = y), equals plus equals are equal, and so on. This is in the setting of equations over the real numbers. We can also try to solve polynomial equations in modular arithmetic, where we consider solutions to be different if they are incongruent. We will focus on the simplest case: a linear congruence ax b mod m. Already in this case we will meet phenomena with no parallel in the case of a real linear equation (Examples 4.2 and 4.3 below). Example 4.. Let s try to solve 8x mod. If there is an answer, it can be represented by one of 0,, 2,..., 0, so we can just run through the possibilities: x mod x mod The only solution is 7 mod : 8 7 = 56 mod. This means 7 and 8 are multiplicative inverses in Z/(). We found an inverse for 8 modulo, and can similarly find multiplicative inverses for every nonzero element of Z/(): x x Check in each case that the product of the numbers in each column is in Z/(). Example 4.2. Find a solution to 8x mod 0. We run through the standard representatives for Z/(0), and find no answer: x x

5 MODULAR ARITHMETIC (SHORT VERSION) 5 In retrospect, we can see a priori why there shouldn t be an answer. If 8x mod 0 for some integer x, then we can lift the congruence up to Z in the form 8x + 0y = for some y Z. But this is absurd: 8x and 0y are even, so the left side is a multiple of 2 but the right side is not. Example 4.3. The linear congruence 6x + 4 mod 5 has three solutions! following table we can see the solutions are 3, 8, and 3: x x In the These examples show us that a linear congruence ax b mod m may not behave like real linear equations: there could be no solutions or multiple solutions. In particular, taking b =, some nonzero elements of Z/(m) may have no multiplicative inverse. The obstruction to inverting 8 in Z/(0) extends to other moduli in the following way. Theorem 4.4. For integers a and m, the following are equivalent: there is a solution x in Z to ax mod m, there are solutions x and y in Z to ax + my =, a and m are relatively prime. Proof. Suppose ax mod m for some x Z. Then m ( ax), so there is some y Z such that my = ax, so ax + my =. This equation implies a and m are relatively prime since any common factor of a and m divides ax + my. Finally, if a and m are relatively prime then by Bezout s identity (a consequence of back-substituting in Euclid s algorithm) we can write ax + my = for some x and y in Z and reducing both sides mod m implies ax mod m. This explains Example 4.2, since 8 and 0 have a common factor of 2. Similarly, we see at a glance that there is no solution to 3x mod 5 (common factor of 3) or 35x mod 77 (common factor of 7). Example 4.5. Since (8, ) =, 8 has a multiplicative inverse in Z/(). We found it by an exhaustive search in Example 4., but now we can do it by a more systematic approach: = 8 + 3, 8 = , 3 = 2 +, so = 3 2 = 3 (8 3 2) = = 3 ( 8) 8 =

6 6 KEITH CONRAD Reducing the equation = modulo, 8( 4) mod. The inverse of 8 in Z/() is 4, or equivalently 7. To summarize: solving for x in the congruence ax mod m is equivalent to solving for integers x and y in the equation ax + my = (be sure you see why!), and the latter equation can be solved without any guesswork by reversing Euclid s algorithm on a and m when (a, m) =. If Euclid s algorithm shows (a, m), then there is no solution. In the real numbers, every nonzero number has a multiplicative inverse. This is not generally true in modular arithmetic: if a 0 mod m it need not follow that we can solve ax mod m. (For instance, 4 0 mod 6 and 4 mod 6 has no multiplicative inverse.) The correct test for invertibility in Z/(m) is (a, m) =, which is generally stronger than a 0 mod m. Although invertibility in Z/(m) is usually not the same as being nonzero in Z/(m), there is an important case when these two ideas agree: m is prime. Corollary 4.6. For a prime number p, an integer a is invertible in Z/(p) if and only if a 0 mod p. Proof. If a mod p is invertible, then (a, p) =, so p does not divide a. For the converse direction, suppose a 0 mod p. We show (a, p) =. Since (a, p) is a (positive) factor of p, and p is prime, (a, p) is either or p. (The proof would break down here if p were not prime.) Since p does not divide a, (a, p) p, so (a, p) =. Therefore the congruence ax mod p has a solution. The upshot of Corollary 4.6 is that our intuition from algebra over R carries over quite well to algebra over Z/(p): every nonzero number has a multiplicative inverse in the system. This is not true of Z/(m) for composite m, and that is why modular arithmetic in a composite modulus requires more care. Why should we care about inverting integers in Z/(m)? (By inverting we always mean inverting multiplicatively. ) One reason is its connection to inverting matrices with entries in Z/(m). Given a square n n matrix A with entries in Z/(m), your experience with linear algebra in R may suggest a matrix with entries in Z/(m) is invertible whenever its determinant is nonzero in Z/(m). But this is false. Example 4.7. We work with matrices having entries in Z/(0). Let A = ( 3 ). The determinant of A is 2 8 mod 0, so det A 0 mod 0. However, there is no inverse for A as a mod 0 matrix. We can see why by contradiction. Suppose there is an inverse matrix, and call it B. Then AB ( 0 0 ) mod 0. (Congruence of matrices means congruence of corresponding matrix entries on both sides.) Writing B = ( x y z t ), we compute AB to get x+3z y+3t ( x+z y+t ) ( 0 0 ) mod 0. Then x + 3z mod 0 and x + z 0 mod 0. The second congruence says x z mod 0, and replacing x with z in the first congruence yields 2z mod 0. But that s absurd: 2z is even and is odd, so 2z mod 0. (Said differently, if 2z mod 0 then 2z = + 0y for some integer y, so 2z 0y =, but the left side is even and is not even.) As a real matrix, A is invertible and A /2 3/2 = ( ). This inverse makes no sense if /2 /2 we try to reduce it modulo 0 (what is /2 mod 0?!?), and that suggests there should be a problem if we try to invert A as a mod 0 matrix.

7 MODULAR ARITHMETIC (SHORT VERSION) 7 Let s look at determinants behave in modular arithmetic. Suppose n n matrices A and B satisfy AB I n mod m. Taking determinants of both sides tells us (by Theorem 3.2) that (det A)(det B) mod m, so det A is invertible in Z/(m). Invertibility of det A in Z/(m) is usually a stronger condition than det A 0 mod m. For instance, the 2 2 matrix A in Example 4.7 has determinant 8 mod 0, which is not invertible. Thus the matrix A is not invertible mod 0. That sure is an easier way to see A is not invertible than the tedious matrix calculations in Example 4.7! Example 4.8. Let m = 4 and A = ( ) as a matrix with entries in Z/(4). The determinant of A is 2 2 = 0 4 mod 4, which is not invertible. Even though A has a nonzero determinant, there is no matrix inverse for A over Z/(4). We now see that we have to be able to recognize invertible elements of Z/(m) before we can recognize invertible matrices over Z/(m), because an invertible matrix will have an invertible determinant. If we want to do linear algebra over Z/(m) then we need Euclid s algorithm (and Bezout s identity) to invert determinants.

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