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Suppose for all $n\in\mathbb Z$, we have $(x + 4n)^2\equiv x^2\bmod m$. Find all $m\in\mathbb N$ for which this is a true statement.

I have no idea how to go about finding m. I tried to use the fact that $(x+4n)^2 - x^2$ should be divisible by m, and then used the well defined-ness of $+$ and $×$ operations to deduce something but I failed.

Parcly Taxel
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childishsadbino
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1 Answers1

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$$\forall n:8nx+16n^2\equiv0\bmod m$$ Since $x$ is a variable, we must have $m\mid8n$ and $m\mid16n^2$. We can ignore the second condition as it is implied by the first one. Since $n$ can be any integer, including 1, we must have $m\mid8$, so $m=1,2,4,8$.

Parcly Taxel
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  • OP needs to clarify what $x$ means, e.g. $x$ isn't a "variable" in $,x^p\equiv x\pmod{p}$. – Bill Dubuque Nov 16 '18 at 04:26
  • @BillDubuque I realize that, but the homework problem I got from my professor didn't mention what x was either, which led to additional confusion. I just assumed x to be fixed and n to be varied, and then repeating the same process with all other x. – childishsadbino Nov 16 '18 at 19:19