2015/086) Let $S=y^2+20y+12$ y is a positive integer.

What is the sum of all possible values of y for which S is a perfect square.

Solution

let $S = x^2$

so $x^2 = (y+10)^2 – 88$

or $88 = (y+10)^2-x^2 = (y+10 + x) (y+10-x)$

both the terms on RHS has to be even as one add and one even shall give fractional x and y

so $(y+10+x) = 44, (y + 10 – x) = 2$ giving $y = 13, x = 21$

or $(y+10+x) = 22, (y+10-x) = 4$ giving $y=3 , x = 9$

so sum of all possible values of $y = 16$