<s>[INST] $3^k$ not congruent to $-1 \pmod {2^e}, e > 2$. $3^k \not\equiv -1 \pmod {2^e}$ for $e > 2, k > 0$. Is this true? I have tried to prove it by expanding $(1 + 2)^k$. [Notation: $(n; m) := n! / (m! (n - m)!)$] E.g., for $e = 3$ I get: $(1+2)^k + 1 = 2 + (k; 1) 2 + (k; 2) 2^2 + (k; e) 2^e + ...$ So, here it's enough to prove that $2^3$ does not divide $2 + (k; 1) 2 + (k; 2) 2^2$. The validity for general e seems very hard to prove. 
 [/INST] Based on pedja,we only prove that $8 \nmid 3^k+1$,
1.when $k=2m+1$,$3^k+1=3^{2m+1}+1=9^m \times 3+1 \equiv 4 \pmod{8} $
2.when $k=2m$,$3^k+1=3^{2m}+1=9^m+1 \equiv 2 \pmod{8} $
so we have $8 \nmid 3^k+1$,this is
$3^k \not\equiv -1 \pmod {2^e}$
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