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.

(In reply to Method by Jer)

Let n=5b.
In the quadratic formula the discriminant becomes (5b)²- 4(5b)b - 4b² = b²
And so a = (5b±b)/2 = {3b,2b}
Which explains all of the observed patterns and shows an easy way of finding a and b for any multiple of 5.

It does not prove there are no solutions where n is not a multiple of 5.

So now suppose n = 2b+c.
The discriminant becomes b²+6bc+c².  The only value of c that makes this a square is 0.

Posted by Jer on 2013-09-04 13:57:20