# Showing that $-1\not\equiv 5^{n}\pmod{2^k}$.

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