250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 37a, b, c give three different remainders upon dividing by 3, put ro = O. If twoor more among a, b, c give the same remainder e upon dividing by 3, putro = I-e. Clearly, at least two of the numbers a+ro, b+ro, c+ro will benot divisible by 3.Now, Jet i denote one of the numbers 1, 2, ... , s. In view of Problem 39(and the fact that qi > 3), there exists an integer r/ such that none of thenumbers a+rh b+rh c+ri is divisible by qt. According to the Chineseremainder theorem, there exist infinitely many positive integers n such thatandn == r (mod 2), n == ro (mod 3),n==ri(modqi)fori=1,2, ... ,s.We shall show that the numbers a+n, b+n and c+n are pairwise relativelyprime. Suppose, for instance, that (a+n, b+n) > 1. Then, there would exista prime q such that q\a+n and qjb+n, hence q\a-b, which implies q\h andh::F ±1. Since n == r (mod 2) and at least two of the numbers a+r, b+r,c+r are odd, at least two of the numbers a+n, b+n, c+n are odd, and wecannot have q = 2. Next, since n == ro (mod 3) and at least two of the numbersa+ro, b+ro, c+ro are not divisible by 3, at least two of the numbers a+n,b+n, c+n are not divisible by 3, and we cannot have q = 3. Since q\h, inview of the definition of h, we have q = qi for a certain i from the sequence1, 2, ... , s. However, in view of n == rj (mod q,), or n == ri (mod q), and inview of the fact that none of the numbers a+rh b+ri, c+ri is divisible byqh none of the numbers a+n, b+n, c+n is divisible by q/ = q, contrary tothe assumption that q\a+n and qJb+n. Thus, we proved that (a+n, b+n)= 1. In a similar way we show that (a+n, c+n) = 1, and (b+n, c+n) = 1.Therefore the numbers a+n, b+n, and c+n are pairwise relatively prime.Since there are infinitely many such numbers n, the proof is complete.46. Such numbers are for instance a = 1, b = 2, c = 3, d = 4. In fact,for odd n, the numbers a+n and c+n are even, hence not relatively prime,and, for even n, the numbers b+n and d+n are even, hence not relativelyprime.47. If n is odd and> 6, then n = 2+(n-2), where n-2 is odd and> 1, and we have (2, n-2) = 1.The following proof for the case of even n > 6 is due to A. Mq,kowski.If n = 4k, where k is an integer > 1 (since n > 6), then n = (2k-1)+