T3MO 2017

For $n\le 6$, we can easily check that $n=1,2,4$ are the only positive integers satisfying the condition.

We will prove that there are no other positive integers which satisfy the condition. Assume to the contrary that there exists a positive integer $n>6$ that satisfies the condition. Let $4n!-4n+1=x^2$ for some non-negative integer $x$.

So $4n!-4n+4=x^2+3$. Let $p$ be a prime factor of $n-1$. From $(n-1)\mid 4n!$ and $(n-1)\mid -4n+4$, we get that $x^2+3$ is divisible by $p$. This means $\left(\frac{-3}{p}\right)=1$ or $p=3$. But from the quadratic reciprocity theorem, $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$ which is 1 only when $p\equiv 1(\mathrm{mod}3)$. So every prime factor of $n-1$ is 3 or is congruent to 1 modulo 3. That is $n-1\equiv 0(\mathrm{mod}3)$.

Case 1 $n-1\equiv 1(\mathrm{mod}3)$; that is $n\equiv 2(\mathrm{mod}3)$ So $4n!-4n+1\equiv 2(\mathrm{mod}3)$, which contradicts the fact that $4n!-4n+1$ is a perfect square.

Case 2 $n-1\equiv 0(\mathrm{mod}3)$; that is $n\equiv 1(\mathrm{mod}3)$ So $4n!-4n+1\equiv 0(\mathrm{mod}3)$ or $3\mid x^2$ or $3\mid x$. Then $9\mid x^2=4n!-4n+1$. But since $n>6$, this implies that $9\mid n!$. So $9\mid -4n+1$ or $n\equiv 7(\mathrm{mod}9)$. Thus $n-1\equiv 6(\mathrm{mod}9)$, which contradicts our conclusion that all prime factors of $n-1$ are 3 or are congruent to 1 modulo 3.

So the only positive integers satisfying the condition are 1, 2 and 4.