A set S of k distinct integers n_{i} is such that ∏ n_{i} divides ∏ (n_{i} + m) for all integers m. Must 1 or -1 belong to S? If all members of S are positive, is S necessarily just {1, 2, ... , k}?

**Solution**

Answer: yes, yes.

m is allowed to be *negative*, so we can use the weaker condition that ∏ n_{i} ≥ ∏ (n_{i} + m). Suppose neither 1 nor -1 is in S. Then taking m = 1 and -1 gives non-zero products. Moreover the product of their absolute values is ∏ |n_{i} + 1| |n_{i} - 1| = ∏ |n_{i}^{2} - 1| < (∏ n_{i})^{2}. So at least one of the absolute values is less than | ∏ n_{i} | (and hence cannot be divisible by it). Contradiction.

Now consider the case where all members of S are positive. Let m be the smallest positive integer not in S. Then ∏ n_{i} = | ∏ (n_{i} - m) |, where we take the product over the members of S less than m. For any member x of S greater than m, we obviously have 0 < x - m < m. So if there are any such members, then ∏ n_{i} > |∏ (n_{i} - m) | (taking the products over all members of S). Contradiction. Hence S must be just 1, 2, ... , m - 1.

Of course, the question becomes more interesting if we restrict m to be positive.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001