lean_workbook_plus_9526

Given $n > 2$ . Prove that : $2^{n} - 1 \ne 3^{q}$ With $q$ is positive integers. Suppose,there exist $2^n=3^q+1$ for $n>2$ . Now, $v_{2}(3^q+1)=v_{2}(4)+v_{2}(q)$ $\Rightarrow v_{2}(2^n)=2+v_{2}(q)$ $\Rightarrow n=2 +v_{2}(q)$ $\Rightarrow n-2=v_{2}(q)$ $\Rightarrow 2^{n-2}\mid q$ . so we assume $q=t.2^{n-2}$ . so, ${(3^{2^{n-2}}})^t +1 > 2^{2^{n-2}} >2^{n-2}$ for n>2 . So,no solution.