Positive integers a and b are such that (15a+16b) and (16a-15b) are perfect squares. Find the least possible value of the smaller of these two squares.

(In reply to Solution - not sure if least by goFish)

Say (16a-15b) is the smaller of the two squares.

Then its least possible value given integral a,b is zero.

This occurs when a=15p and b=16p for integral p.

Then (15a+16b) = 481p.

Since 481 is square-free, p=481 is the least solution resulting in a square.

So a=7215, b=7696 gives (15a+16b)=481^2 and (16a-15b)=0.

Posted by xdog on 2005-11-09 15:16:04