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More specifically I want to prove for any $n\geq 1$, $11\nmid 64^n +22^n -2$, and I'm having doubts as to the validity of my proof.

So, if $11$ did indeed divide $64^n +22^n -2$ for any $n\geq 1$, then, working modulo 11 would yield $$ \bar{64}^n +\bar{22}^n -\bar{2} = \bar{0} \text{ (mod11)} $$ but notice that $\bar{64}=\bar{-2}$ and $\bar{22}=\bar{0}$. Thus, $\bar{(-2)}^n +\bar{0}^n -\bar{2} = \bar{(-2)}^n - \bar{2} = \bar{0}$. Computing a few values of $\bar{(-2)}^n$, we find, $\bar{-2}, \bar{4}, \bar{8}, \bar{5}, \bar{1}, \bar{-2}$, and so more specifically, $\bar{(-2)}^n$ is periodic with period 5 and never equal to $\bar{2}$ within that period, so the equality $\bar{(-2)}^n - \bar{2} = \bar{0}$ can never hold for any $n\geq1$.

Let me know if this proof is sufficient or if I need to add or change anything, any help would be appreciated.

prill
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  • looks good to me – LurchiDerLurch Dec 06 '21 at 22:40
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    Your proof looks fine. If (or when) you know about quadratic residues, an alternative is to note that $\left(2\over11\right)=\left(-9\over11\right)=\left(-1\over11\right)=-1$ (since $11\equiv3$ mod $4$). – Barry Cipra Dec 06 '21 at 22:47
  • bug: $,(-2)^3\equiv -8$ (not $8).,$ Luckily $,-8\not\equiv 2,$ so the proof still works. – Bill Dubuque Dec 07 '21 at 18:27
  • @BarryCipra Another argument for $(\frac{2}{11})=-1$ is that $(\frac{2}{p})=1$ for a prime $p$ holds if and only if $p$ is of the form $8k\pm1$ , which $11$ is not. – Peter Dec 08 '21 at 10:14
  • Generally it's easier to use Euler's Criterion as here in the linked dupe, i.e. $\bmod 11!:\ 2\equiv 8^{2n}\overset{(\ \ )^{\Large 5}}\Longrightarrow -1\equiv (8^n)^{10}\equiv 1,$ contra lil Fermat. The comments using reciprocity are essentially doing the same thing. – Bill Dubuque Feb 03 '22 at 11:59

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