Math 409/409G History of Mathematics

1. Math 409/409GHistory of Mathematics The Theory of Congruences

2. What are congruences? If you are familiar with modular or clock arithmetic, then you already know what congruences are. Definition: Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n if n divides a – b. Notation: a≡ b (mod n) or simply a≡nb.

3. Examples: 24 ≡7 3 since 7│(24 – 3) 3 ≡7 24 since 7│(3 – 24) -31 ≡7 11 since 7│(-31 – 11) since 7 (25 – 12) Theorem: a ≡n b if and only if a and b have the same remainder when divided by n. Proof: First assume that a and b have the same remainder when divided by n. Then there are integers q and Q such that a qn + r and b  Qn + r. Thus a ≡n b since a – b (q – Q)n.

4. Now assume that a ≡n b. Then a – b  ns for some integer s. And by the division algorithm there exist integers q, Q, r, and R such that a qn + r and b  Qn + R with 0  r, R < n. So But –n <r - R < n. So n│(r – R) implies that r – R 0. Thus a and b have the same remainder when divided by n.

5. Since adding a multiple of n to a number does not change the remainder when that number is divided by n, we have: Corollary:a≡na + kn for any integer n. Example:-31 ≡7 -31 + 5·7 = 4. So the remainder when -31 is divided by 7 is 4. The same result could, of course, been achieved by doing a long division problem.

6. Properties of Congruences For integers a, b, c, d and positive integers n, k: • a≡na. • a≡nb b≡na. • a≡nb and b≡nc  a≡nc. • a≡nb and c≡nd  a + c≡nb + d and ac≡nbd. • a≡nb a + c≡nb + c and ac≡nbc. • a≡nb ak≡nbk.

7. Proof thata≡nb and c≡nd  ac≡nbd

8. Proof that a≡nb  ac≡nbc This follows from the previous property. That is, a≡nb and c≡nc  ac≡nbc. Or one could give a direct proof.

9. Proof that a≡nb  ak≡nbk This statement is proved by induction. Clearly the statement holds for k 1. Assume that it is true for some fixed k. Then So by induction, the statement is true for all positive integers k.

10. Is 220 – 1 divisible by 41? Answering this question is equivalent to finding the remainder when 220 – 1 is divided by 41. Using congruences makes this easy. Find the remainder when 220 is divided by 41 by first finding the remainder when smaller powers of 2 are divided by 41. Then you can use the property a≡nb  ak≡nbk to find the remainder for larger powers of 2.

11. You may be tempted to use 26 since this is the smallest power of 2 that is larger than 41. But 26 64 ≡41 23 and 23 is a rather large number to raise to power when using the property a≡nb  ak≡nbk to find the remainder for larger powers of 2. So use 25 since 25 32  41 – 9 ≡41 -9 and it’s easier to find powers of -9 than powers of 23.

12. Is 220 – 1 divisible by 41? So 220 – 1 ≡41 1 – 1  0. Since the remainder when 220 – 1 is divided by 41 is 0, the answer is yes, 220 – 1 is divisible by 41.

13. What’s the remainder when N  1! +2! + 3! + ··· + 100!is divided by 4? When n 4, 4 appears in the product of n!. So n! ≡4 0 when n 4. This gives 1! +2! + 3! + ··· + 100! ≡41! +2! + 3! ≡4 3. So the remainder when N is divided by 4 is 3.

14. This ends the lesson on The Theory of Congruences