As the title says, I'm trying to prove that $-1\not\equiv 5^{n}\pmod{2^k}$.

So far I'm trying to proceed by contradiction. I've assumed that $-1\equiv 5^{n}\pmod{2^k}$, so that there exists an integer $m$ such that $-1=5^{n}+m\cdot2^{k}$ so that $5^{n}+1=m\cdot 2^{k}$.

So we have shown that $2^{k}|5^{n}+1$. Working out a few examples this seems like it may be a contradiction, but I can't see why in the general case. Clearly $2^{k}$ does not divide $5^{n}$, but that pesky $1$ is getting in the way!

I feel like this whole problem should be really simple, but just can't seem to get it. Any help would be mega appreciated.


Presumably you mean $k\ge 2$. Since $5^n\equiv 1^n\equiv1\pmod 4,\,$ $5^n+1$ can't be divisible by $4$.

  • 1
    $\begingroup$ Thanks, I did mean to specify $k \geq 2$. $\endgroup$ – CoffeeCrow Feb 4 at 11:55

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