Perfect Square Factorials

Given any prime number p, there's another prime number between p and 2p, per Bertrand's Postulate.

By Bertrand's Postulate, for any positive integer $n\ge 2$ , there is a prime $q$ such that $\frac{n}{2}<q \le n$ . Thus, $q$ divides $n!$ . Now for $n \ge 5$ , $q^2 > n$ , so if $n!$ has unique prime factorization $p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r}$ , then the exponent of $q$ must be $1$ . Since perfect squares must have even exponents in their prime factorizations, we know $n!$ cannot be a perfect square for $n \ge 5$ . Checking $n=0, 1, 2, 3,$ and $4$ , shows only $n=0, 1$ are perfect squares. So there are only $2$ solutions.

Here is a logical solution. For a number to be a perfect square, power of each of its prime factor must be even. But in n! , power of the largest prime p(<n) is always 1, thus for all n(>1) , n! is never a perfect square. So answer is 2.

Bertrand's postulate(in actuality a theorem) states that for every prime p, there exists another prime number between p and 2p. This means that ∀n>1,n! will always have a single power of some prime number(s).

For any positive integer n≥2, there exists a prime p such that n/2<p≤n This implies that p∣n!. For n≥5,p2>n. So if n! has unique prime factorization (p1)^k1⋅(p2)^k2⋯(pr)^kr, then the exponent of p must be 1. Since perfect squares must have even exponents in their prime factorizations, we know n! cannot be a perfect square for n≥5.

So,we have 0! and 1! which are perfect squares (1)^2 and hence the solution is 2.