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When is $n^2 - 8056$ a perfect square ? What is a generalized solution for $n^2 - c$ ?

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    $\begingroup$ What have you tried? $\endgroup$
    – Arthur
    Aug 20, 2019 at 17:37

1 Answer 1

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Hint: $n^2 - 8056 = m^2$ iff $(n-m)(n+m)= 8056$. This equation has integer solutions iff $8056$ can be decomposed into two factors of the same parity.

This argument holds in general: an integer is a difference of two squares iff it is odd or a multiple of $4$.

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  • $\begingroup$ See also math.stackexchange.com/a/547420/589 $\endgroup$
    – lhf
    Aug 20, 2019 at 18:03
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    $\begingroup$ Nice. And this gives explicit solutions for $n$ too (after factoring of course). Write $m$ here $m=8056 = xy$; $x$ and $y$ positive integers both $x$ and $y$ even ($m$ a multiple of 4) or both $x$ and $y$ odd $(m$ odd). Then $n=\frac{x+y}{2}$. $\endgroup$
    – Mike
    Aug 20, 2019 at 20:28

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