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 $p \equiv 1 \pmod 3$. So every prime factor of $n-1$ is 3 or is congruent to 1 modulo 3. That is $n-1 \equiv 0 \pmod 3$.

Case 1 $n-1 \equiv 1 \pmod 3$; that is $n \equiv 2 \pmod 3$
So $4n! - 4n + 1 \equiv 2 \pmod 3$, which contradicts the fact that $4n! - 4n + 1$ is a perfect square.

Case 2 $n-1 \equiv 0 \pmod 3$; that is $n \equiv 1 \pmod 3$
So $4n! - 4n + 1 \equiv 0 \pmod 3$ 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 $n \equiv 7 \pmod 9$. Thus $n-1 \equiv 6 \pmod 9$, 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.