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For every natural number $n$, prove that $4\mid(3^n-1)$ iff $4$ does not divide $3^n+1$.

Since it's a bi-conditional it has to be proven both ways.

It's easy to see if $4\mid(3^n-1)$ then $3^n-1=4x$ where $x$ is an integer then $3^n=4x+1$. We can plug this into $3^n+1$, which gives $(4x+1)+1=4x+2$ which is not divisible by $4$ because it'll always have a remainder of $2$.

I don't know how to prove the inverse implication "if $4$ does not divide $3^n+1$ then $4\mid(3^n-1)$". I thought maybe if $4$ does not divide $3^n+1$ then it can be defined as $4k+i$ where $i \in \{1,2,3\}$, but I tried plugging that into $3^n-1$ and it didn't work out.

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    $\begingroup$ That $3^n$ is a red herring. Try this: "If $k$ is an odd number, then $4\mid (k-1)$ iff $4$ does not divide $k+1$." $\endgroup$ – TonyK Oct 1 at 9:14
  • $\begingroup$ Hint $ $ Viewed $\bmod 4\,$ it is: $\,(-1)^{\large n}\equiv 1\iff (-1)^{\large n}\not\equiv -1,\, $ which is true for any modulus $\neq 2\ $ $\qquad\ \ \ \ \ \ \ \ $ $\endgroup$ – Bill Dubuque Oct 1 at 14:21
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Note that both numbers $3^n-1$ and $3^n+1$ are even numbers and that their difference is $2$. Therefore, one and only of them is a multiple of $4$.

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Given \begin{align}3^n+1\not\equiv0\pmod4&\implies(-1)^n\not\equiv-1\pmod4\\&\implies(-1)^n\equiv1\pmod4,\end{align} we have that $$3^n-1\equiv(-1)^n-1\equiv1-1\equiv0\pmod4$$ as desired.

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  • $\begingroup$ Sorry how is $3^n+1$ (mod4) go to $(-1)^n$ is congruent to 1(mod4)? $\endgroup$ – Dylan Y Oct 2 at 1:30
  • $\begingroup$ Because $3\equiv-1\pmod4$ so $3^n\equiv(-1)^n\pmod4$. Now we know that $(-1)^n$ can only take the values $-1$ and $1$, but we are given that $3^n+1\not\equiv0\pmod4\implies(-1)^n\not\equiv-1\pmod4$. Therefore $(-1)^n\equiv1\pmod4$. $\endgroup$ – TheSimpliFire Oct 2 at 6:35

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