By Richard Bellman

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**Example text**

Taking m = 37 and n = 15 gives (A, B, C) = (154, 231, 385) and d = 77. 6. Let a, b, and c be integers. Prove that gcd(a, b) gcd(b, c) gcd(c, a) gcd(a, b, c)2 and lcm(a, b) lcm(b, c) lcm(c, a) lcm(a, b, c)2 are equal integers. β γ β γ Solution. Let a = p1α1 · · · pnαn , b = p1 1 · · · pn n , and c = p1 1 · · · pn n , where p1 , . . , pn are distinct primes, and α1 , . . , αn , β1 , . . , βn , γ1 , . . , γn are nonnegative integers. Then n gcd(a, b) gcd(b, c) gcd(c, a) = gcd(a, b, c)2 min{αi ,βi } n min{βi ,γi } pi i=1 n pi i=1 n min{γi ,αi } pi i=1 2 min{αi ,βi ,γi } pi i=1 n min{αi ,βi }+min{βi ,γi }+min{γi ,αi }−2 min{αi ,βi ,γi } = pi i=1 and n lcm(a, b) lcm(b, c) lcm(c, a) = lcm(a, b, c)2 max{αi ,βi } n i=1 n max{βi ,γi } pi pi i=1 n i=1 2 max{αi ,βi ,γi } pi i=1 n = max{αi ,βi }+max{βi ,γi }+max{γi ,αi }−2 max{αi ,βi ,γi } pi i=1 max{γi ,αi } pi .

Let d be the greatest common divisor of k and p. Then k = k1 d, p = p1 d, for some integers p1 , k1 with gcd( p1 , k1 ) = 1. Hence k1 a = p1 b, and since gcd(a, b) = 1, we have k1 = b, p1 = a. This is false, because k1 < b. It follows that one of the numbers from (1) has the remainder 1 when divided by b, so there is u ∈ {1, 2, . . , b − 1} such that au = bv + 1 and the lemma is proved. In fact, there are infinitely many positive integers u and v satisfying this property, that is, u = u 0 +kb, v = v0 +ka, where u 0 and v0 satisfy au 0 −bv0 = 1.

M k are pairwise relatively prime, then a ≡ b (mod m i ), i = 1, . . , k, if and only if a ≡ b (mod m 1 , . . , m k ). 30 I Fundamentals, 1. Divisibility Let us prove the last property. From a ≡ b (mod m i ), i = 1, . . , k, it follows that m i | a − b, i = 1, . . , k. Hence a − b is a common multiple of m 1 , . . , m k , and so lcm(m 1 , . . , m k ) | a − b. That is, a ≡ b (mod lcm(m 1 , . . , m k )). Conversely, from a ≡ b (mod lcm(m 1 , . . , m k )), and the fact that each m i divides lcm(m 1 , .