25 Therefore, 1 is a linear combination of rr0 and m ; as 1 is obviously the smallest positive linear combination, it must be their gcd. 1.59 Let a , b and d be integers. If d = sa + tb , where s and t are integers, f nd in f nitely many pairs of integers ( s k , t k ) with d = s k a + t k b . Solution. If d = sa + tb , and if we de f ne s k = s + kb and t k = t − ka , then d = s k a + t k b for all k . 1.60 If a and b are relatively prime and if each divides an integer n , prove that their product ab also divides n . Solution. Assume that ( a , b ) = 1 and n = ak = b ` . By Corollary 1.40, b | ak implies b | k . Thus, k = bk0 and so n = ak = abk0 . 1.61 Prove, for any (possibly negative) integers a and b , that ( b , a ) = ( b − a , a ) . Solution. If c is a common divisor of a and b , then c | a and c | b ; hence, c | b − a , and c is a common divisor of a and b − a . This does not yet show that the two gcd ’ s are equal. However, if c0 is a common divisor of a and b − a , then
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Greatest common divisor, Euclidean algorithm, common divisor