SquareNumber Basic

Problem - 2391
Find such a positive integer $n$ such that both $(n+23)$ and $(n-30)$ are square numbers.

Answer     706

Let $n+23= m^2$ and $n-30=k^2$, then we have $$(n+23)-(n-30)=m^2 - k^2 \implies 53 = (m+k)(m-k)$$

Because $53$ is a prime number which can only be factorized as $1\times 53$, Hence we must have $$m-k = 1, m+k=53\implies k=26, m=27$$

It follows that $n=m^2 -23 = \boxed{706}$.

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