Find all n such that $3^{(2n-1)}-2^{(n-1)}$ is perfect square 
Find all n (natural number) such that $3^{(2n-1)}-2^{(n-1)}$ is perfect square. 


For $n \ge 4$ , we have $2^{n-1} \equiv 0\mod8$ while $3^{2n-1} \equiv 3 \mod 8 $ $($ Since $2n-1$ is always odd $)$. So 
$$3^{(2n-1)}-2^{(n-1)} \equiv 3\mod 8$$
But for a perfect square , we have $k^2 \equiv 0,1,4 \mod 8$.
So no solution occurs for $n\ge4$.
Now , we can simply check for all values between $1,3$ and find that the only solution is at $n=2.$
