# Maximize value of (a+b) such that (a*a-b*b = N).

Given a value N, find the maximize value of value of (a+b) such that a^2-b^2 = N

Note: The value of N is odd. a and b are integers.

For eg if n = 1

The value of aa - bb is 1. The maximum value occurs when a=1 and b=0.

Thus, a+b = 1.

I know it's anser will always be n.

But can anyone help me in knowing that how it answer will always be n.

Note that $$a \neq b$$. $$a+b=\frac N {a-b}$$ Since $$|a-b| \geq 1$$ we get $$a+b \leq N$$. The value $$N$$ is attained when $$a=\frac {N+1} 2$$ and $$b=\frac {N-1} 2$$