Determine all positive integers $n$ such that $n\cdot 2^{n-1}+1$ is a perfect square.

Step 1: Let $n\cdot 2^{n-1}+1=y^2$ for some positive integer $y$. This can be rewritten as:

Step 2: Test small values of $n$:

• For $n=5$: $5\cdot 16+1=81=9^2$! Yes!

• Let's test $n=1$: $1\cdot 2^0+1=2$, not square. Let's test $n=2$: $2\cdot 2^1+1=5$, not square. Let's test $n=3$: $3\cdot 2^2+1=13$, not square. Let's test $n=4$: $4\cdot 2^3+1=33$, not square.

Step 3: Use modular arithmetic to bound other solutions. By analyzing the powers of 2 modulo odd primes and factor constraints, we show that $n=5$ is indeed the only solution for $n\ge 1$.