Syler wrote: Find all integers n for which number $ \log_{2n - 1}(n^2 + 2)$ is rational. I consider $ \log_{2n-1}$ is only defined for $ n\ge 2$ and so $ n=1$ is not a solution. The equation is equivalent to $ (n^2+2)^a=(2n-1)^b$ for some positive integers $ a,b$ So, if $ p$ is a prime divisor of $ n^2+2$, it must divide $ 2n-1$, so $ n^2+2+2n-1=(n+1)^2$, so $ n+1$, so $ 2n-1-(n+1)=n-2$, so $ (n+1)-(n-2)=3$ So $ n^2+2=3^c$ and $ 2n-1=3^d$ But $ 2n-1=3^d$ $ \implies$ $ n^2+2=\frac{3^{2d}+2\cdot 3^d+9}4$ and so $ 4\cdot 3^c=3^{2d}+2\cdot 3^d+9$ $ d=0$ $ \implies$ $ c=1$ and $ n=1$, which is not a solution $ d=1$ $ \implies$ $ 3^c=6$, not a solution $ d=2$ $ \implies$ $ c=3$ and the solution $ n=5$ $ d>2$ $ \implies$ $ 4\cdot 3^{c-2}=3^{2d-2}+2\cdot 3^{d-2}+1$ and no solution ($ c=2$ gives no solution and $ c>2$ implies $ 3|LHS$ and not $ 3|RHS$) Hence the unique solution $ \boxed{n=5}$ with $ \log_9(27)=\frac 32$