Four positive integers have the property that the square of the sum of any two is divisible by the product of the other two. Show that at least three of the integers are equal.
Solution
Suppose that we can find 4 such integers with no three equal. Let a, b, c, d be the set with the smallest a + b + c + d. Suppose a prime p divides a and b. Since a divides (b+c)2 and (c+d)2, it follows that p must divide b+c and c+d. Hence it divides c and d. But now a/p, b/p, c/p, d/p has the same property and smaller sum. Contradiction.
Now suppose an odd prime p divides a. Then p must divide b+c, c+d and d+b and hence also their sum 2(b+c+d). But p is odd, so it must divide b+c+d and hence b = (b+c+d) - (c+d). Contradiction. So a must be a power of 2. Similarly, b, c, d must be powers of 2. If two of them are > 1, then they have a common factor 2. Contradiction. So three of them must be 1. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
4 March 2004
Last corrected/updated 4 Mar 04