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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.

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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 $b=\frac {N-1} 2$

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