Determine all pairs (a,b) of positive integers with a > b, for which (a²+b²)/(a-b) is an integer which divides 1995.

Prove that there are no others.

Let n be a factor of 1995
(a²+b²)/(a-b) = n
(a²+b²) = n*(a-b)
a²-na+ b²+nb = 0
which is quadratic in a so
a=[n±√(n²-4(b²+nb))]/2

For each value of n there are only a few values of b that yield real solution so from there I went brute force and got the same list as Charlie.

Note: Only multiples of 5 seem to have solutions.  I plan to look a little deeper.

Posted by Jer on 2013-09-04 13:28:20