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### Congruences2

Course: MATH 461, Fall 2009
School: UNC Asheville
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Word Count: 698

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461, MATH Sections 1 and 2 Abstract Algebra I Fall 2008 Handout 5: cool tricks using congruence arithmetic Let's look at some neat facts about congruence arithmetic, shall we? Proposition. If a Z, then a2 is congruent to one of 0, 1, or 4 modulo 8. Proof. You can show this by using the fact that every integer is congruent to its remainder upon division by 8...so what then? Application. Using the fact that 1000...

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461, MATH Sections 1 and 2 Abstract Algebra I Fall 2008 Handout 5: cool tricks using congruence arithmetic Let's look at some neat facts about congruence arithmetic, shall we? Proposition. If a Z, then a2 is congruent to one of 0, 1, or 4 modulo 8. Proof. You can show this by using the fact that every integer is congruent to its remainder upon division by 8...so what then? Application. Using the fact that 1000 = 8 125 0 mod 8, come up with a 25-digit number that is definitely not a perfect square. Deeper results still are provable: Dirichlet's Theorem on the Distribution of Primes. Let a and b be relatively prime. Then the arithmetic progression {a + bn | n Z} contains infinitely many primes. Moreover, if (b) represents the 1 number of integers x such that 1 x b and (b, x) = 1, then in the limit (b) of all primes lie in the above progression. The general version of this theorem is very hard to prove, but we can easily prove the special case Proposition. There are infinitely many primes p such that p 2 mod 3. Proof. Suppose BWOC there are only finitely many such primes, and let them be p1 , ..., ps . Define s m = 1 + i=1 p2 . You should now be able to compute the congruence class of m modulo 3 in two i different ways, giving a contradiction. Here's some space: Here's a neat little unrelated trick: Algorithm. A positive integer n is divisible by 3 if and only if the sum of its decimal digits is divisible by 3. The same trick works for 9. Proof. Write out the number in decimal digits: n = dk dk-1 d1 d0 . What does this mean? Proposition. If p is prime and a, b Z, then (a + b)p ap + bp mod p. Proof. What does the binomial theorem give? And what do you know from an earlier section about p vis-`-vis the binomial coefficients p ? a i Hands down one of the most useful results in all of number theory is k Fermat's Little Theorem. If p is prime, then ap a mod p for all a Z. More generally, ap a mod p for all Z a and integers k 1. Proof. For the first statement, try induction on a 0, expanding (a + 1)p modulo p in the inductive step. (Note that the second statement, regarding arbitrary k, is another induction!) Let us end by relearning how to solve basic algebraic equations. Theorem. If (a, m) = 1 then for every b Z the congruence ax b mod m has a solution. In fact, x = sb, for any s satisfying sa 1 mod m. From this it follows that any two solutions are congruent modulo m. Proof. First, explain why there is an integer s satisfying as 1 mod m. Then show that sb is really a solution. Finally, show that if x and y are both solutions, they're congruent modulo m. The nice thing about primes is that (a, p) = 1 always holds unless p|a. We thus get a nice... ...Corollary. Suppose that p is prime and p |a. Then the congruence ax b modp is always solvable. Finally, we have a theorem that can be generalized to arbitrarily many congruences whose moduli are pairwise relatively prime: The Chinese Remainder Theorem. If (m, m ) = 1, the set of simultaneous congruences x b mod m x b mod m and has a solution, and all solutions are congruent modulo mm . Proof. T...

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