SquareNumber Basic

Problem - 2390
Find such a positive integer $n$ such that both $(n-100)$ and $(n-63)$ are square numbers.

Answer     424

Let $n-100= m^2$ and $n-63=k^2$, then we have $$(n-63)-(n-100)=k^2 - m^2 \implies 37 = (k+m)(k-m)$$

Because $37$ is a prime number which can only be factorized as $1\times 37$, Hence we must have $$k-m = 1, k+m=37\implies k=19, m=18$$

It follows that $n=m^2 + 100 = \boxed{424}$.

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