By Santos D.A.

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But then 2t − 2s = 96 = 3 · 25 or 2s (2t−s − 1) = 3 · 25 . By unique factorisation, s = 5,t − s = 2, giving s + t = n = 12. 125 Example Prove that in any set of 33 distinct integers with prime factors amongst {5, 7, 11, 13, 23}, there must be two whose product is a square. Solution: Any number in our set is going to have the form 5a 7b 11c 13d 23 f . Thus to each number in the set, we associate a vector (a, b, c, d, f ). These vectors come in 32 different flavours, according to the parity of the components.

N! n! − n!! = = = mod (n − 1). (1 − 1/2! + · · · + (−1)n−1 /(n − 1)! ) (n − 1) m + (−1)n−1n/(n − 1) + (−1)n/(n − 1) (n − 1) (m + (−1)n) , where M is an integer, since (n − 2)! , k ≤ n − 2. 82 Example Prove that 6n+2 k=0 6n + 2 k 3 ≡ 0, 23n+1, −23n+1 2k mod 23n+2 when n is of the form 2k, 4k + 3 or 4k + 1 respectively. Solution: Using the Binomial Theorem, 3n+1 2S := 2 k=0 √ √ 6n + 2 k 3 = (1 + 3)6n+2 + (1 − 3)6n+2 . 2k √ √ Also, if n is odd, with a = 2 + 3, b = 2 − 3, 1 3n+1 (a + b3n+1) 2 = 3n + 1 2 3n + 1 3n+1−2r r 2 3.

Assume that t|a,t|b. Then a = tm, b = tn for integers m, n. Hence d = ax0 + bx0 = t(mx0 + ny0 ), that is, t|d. The theorem is thus proved. ❑ ☞ It is clear that any linear combination of a, b is divisible by (a, b). 89 Lemma (Euclid’s Lemma) If a|bc and if (a, b) = 1, then a|c. Proof: As (a, b) = 1, by the Bachet-Bezout Theorem, there are integers x, y with ax + by = 1. Since a|bc, there is an integer s with as = bc. Then c = c · 1 = cax + cby = cax + asy. ❑ 34 GCD and LCM 35 90 Theorem If (a, b) = d, then a b ( , ) = 1.