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

JohnColtraneisJC
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Tanmay Gajapati
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1 Answers1

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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 $4$.

lhf
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  • See also https://math.stackexchange.com/a/547420/589 – lhf Aug 20 '19 at 18:03
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    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 $n=\frac{x+y}{2}$. – Mike Aug 20 '19 at 20:28