Problem 3

Let $3^m\Vert n$. Expand $(3-1)^n+1$. Since $n$ is odd, the constant terms cancel and the first term is $3n$, whose 3-adic valuation is $m+1$. For a term $\binom nb3^b$ with $b\ge3$, if $b\ge m+2$ it is immediately divisible by $3^{m+2}$. If $3\nmid b$, the product formula for $\binom nb$ shows $v_3(\binom nb)\ge m$, so the term has valuation at least $m+3$. If $3\mid b$, the same product formula gives $v_3(\binom nb)\ge m-v_3(b)$, and $b-v_3(b)\ge2$, so the term has valuation at least $m+2$. Thus $v_3(2^n+1)=m+1$. Since $n^2\mid2^n+1$, we have $2m\le m+1$, hence $m=1$.