For n > 1 is there any number for which n! is a perfect square?

Question

The sequence below contains terms that doesn't have any perfect square factors (except 1).
2, 3, 5, 6, 7, 10, 11, 13, 14, 15...

What is the sum of the square-roots of the terms starting from 2 to the largest two-digit term?

Answer 1

The answer is no

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If n is prime, then its obvious that n! is not a perfect square, for in that case n∣n! but n2∤n!. If n is not a prime, then let the r'th prime pr be the largest prime which is less than n. Now, n!=1⋅2⋅3⋅…⋅pr⋅(pr+1)⋅(pr+2)⋅…⋅(pr+n−pr). We can show that pr does not divide any of the numbers pr+1,pr+2,…,n. This is because any number greater than pr which is divisible by pr must be of the form mpr where m≥2. But by Bertrand's postulate, pr+1<2pr, where pr+1 is the (r+1)th prime. Thus if any of the numbers pr+1,pr+2,…,n is divisible by pr then one of these numbers must be the (r+1)th prime. But this contradicts the fact that pr is the largest prime which is less than n. So none of the numbers pr+1,pr+2,…,n is divisible by pr. So, we see that pr∣n! but p2r∤n!. Hence, n! is not a perfect square. This also proves that n! actually has even number of divisors.