(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 20130904 13:57:20 