<s>[INST] $5^{2^{n-3}}\not \equiv -1 \mod 2^n$ for any $n\geq 3$ I am in the middle of solving an abstract algebra problem and I have narrowed it down to showing that
$$5^{2^{n-3}}\not \equiv -1 \mod 2^n \quad \text{ for any } \quad n\geq 3. $$
I am not very good with congruences. I know this is equivalent to showing that $5^{2^{n-3}}+1\not \equiv 0\mod 2^n$, i.e., $5^{2^{n-3}}$ is not a multiple of $2^n$, but I am still not sure how to prove this.
Thank you for your help.
 [/INST] prove $5^{2^{n-3}}\not\equiv -1$ mod $2^{n}$, $n\ge 3$
for $n=3$:$\space 2^3=8\nmid 5^{2^0}+1=6$
for $n=4$:$\space 2^4=16\nmid 5^{2^1}+1=26$
for $n\ge 5$: let $m=n-3$:$\space$show that $5^{2^m}\not\equiv -1$ mod $2^{m+3}$, $m\ge 2$
that is, show $2^m2^3\nmid 5^{2^m}+1$
$2^m\nmid 5^{2^m}+1\Longrightarrow 2^m2^3\nmid 5^{2^m}+1$
$5^{2^m}+1=(4+1)^{2^m}+1=\sum_{k=0}^{2^m}{{2^m}\choose k}4^k1^{2^m-k}+1\space$ per binomial theorem.
$=1+{{2^m}\choose 1}4+{{2^m}\choose 2}4^2+\dots +{{2^m}\choose {2^m-1}}4^{2^m-1}+4^{2^m}+1$
$=1+\underbrace{{{2^m}\choose 1}4+{{2^m}\choose 2}4^2+\dots +{{2^m}\choose {2^m-1}}4^{2^m-1}+2^{2^m}2^{2^m}}_{divisible\space by\space 2^m}+1$
$=2^mk+2$ which is not divisible by $2^m$ for $m\gt 1$
therefore $5^{2^m}\not\equiv -1$ mod $2^{m+3}$, $m\ge 2$ and thus $5^{2^{n-3}}\not\equiv -1$ mod $2^{n}$, $n\ge 3$
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