Suppose two integers, m and n differ by 1. Suppose also that x, y, and a are integers such that x + a = m^2 and y + a = n^2. Show that xy + a is a perfect square.

without loss of generality as it is symmetric let m = n+ 1 
x=(n+1)^2-a ;
y=n^2 -a

xy + a = ( (n+1)^2 - a)( n^2 - a) +a
= (n^2)(n+1)^2 - (2) (n)(n+1) (a) + a^2
= ((n)(n+1) - a)^2

therefore xy+a is a perfect square.