Smallest Square Factorial

For a number to be a square, its prime factors must each appear an even number of times. So, we’ll examine the behavior of prime factors as we increase $n$ .

We start with $n=2$ , where we have an unpaired factor of 2. The next number with a factor of 2 is twice 2, or 4, so we have to make it to at least $n=4$ to get a square. (Note that 4 has 2 factors of 2, so it actually wouldn’t balance our 2’s, but this turns out not to be important.) Before we can make it to 4, we have another unbalanced prime, 3. The next number with a factor of 3 is twice three, or 6. So, $n$ must be at least 6. You can probably see the pattern: before reaching 6, we get an unbalanced factor of 5. And before reaching 10 (twice 5), we encounter 7.

What we are observing is called Bertrand’s Postulate , which tells us that for every $n>1$ , there exists a prime $p$ such that $n<p<2n$ . In other words, before we can balance the current prime, by reaching its double, we will encounter at least one new, unbalanced prime. So, our square-finding efforts were doomed fron the start, and the primes never balance. The answer is then $\boxed{-1}$

This is almost the same as Jason's solution but structured differently.

First notice that in the prime factorisation of a perfect square, every prime factor occurs (at least) twice.

Then notice that highest prime factor of n! (n>1) occurs exactly once! ( If p is the highest prime factor of n! we know by Bertrand's postulate that 2p (and so all larger multiples of p) are more than n and so do not appear in n! Furthermore a second p cannot be made by multiplying other numbers from n!, because p is a prime.)

Put the two observations together and we are done.

(Thanks to those who pointed out that I cannot avoid appealing to Bertrand to complete this solution)

Starting the product with 2 3 4*….., we note that when we come to a prime p, we must extend the product to 2p in order for the product to have the factor p^2. But there is always a new prime, q, such that p < q < 2p, so we must extend the product to 2q. The argument can be extended forever by Bertrand's postulate, so there cannot be any perfect square. Ed Gray

I approached it the same way, observing that all we need to show that there aren’t any perfect square factorials is to show that there is always at least one prime greater than n/2 and less than or equal to n. I didn’t know Bertand’s Postulate, cited by Jason Carrier, but I did look at the number of primes between n/2 and n as n increases exponentially (100, 1000, etc.) and the number of primes does continue to increase so there’s always at least one!

If $n! = n(n-1) \cdots 2 = k^2$ for some $k \in \mathbb{N}$ , then every prime $p \leq n$ divides $k^2$ , and consequently, every such prime $p \mid k$ . If $q$ is the largest prime $\leq n$ , then $q^2 \nmid n!$ . Otherwise, $\exists r \in \{q+1, q+2, \cdots, n\}$ that is a multiple of $q$ , implying $n \geq 2q$ . Since for any positive integer $x$ , there is a prime between $x$ and $2x$ , there is a prime $s$ such that $q < s < 2q \leq n$ contradicting the maximality of $q$ . Since $q^2 \mid k^2$ yet $q^2 \nmid n!$ , $n! \neq k^2$ .

Perfect squares are integers that have two equivalent factors, however more formally see Jason Carrier’s answer.

The number 16 has the two factors 4, 81 has two 9s and so on.

Factorials however multiply a number by all the positive integers preceding and it, so there are no two identical factors to make it a perfect square.

It's not very rigorous, but here goes. Every n! beyond n>1 will contain a 2 as a factor. If you take the square root of that product n!, then you will get the square root of a 2 as a part of that product. With the square root of a 2 being irrational, every product will be irrational and therefore can't be a perfect square.

In factorial we don’t twice numbers but in perfects squares we do twice numbers

Let $p$ be the greatest prime in the range $1$ to $n$ inclusive. Because of Bertrand's Postulate, which states that there is always a prime between an integer $k$ and $2k$ , $n < 2p$ otherwise it would contradict the fact that $p$ is the greatest prime in the range. This further implies that $n < 2p \leq p^2$ (Equality occurs when $n=2$ ). For a number to be a square number, its prime factorization has to contain primes all of which are elevated to an even power. However in the prime factorization of $n!$ , $p$ will be to the power of $1$ , which is not even, and hence $n!$ cannot be a square.

For n! As n keeps on increasing your self with an uneven amount of factors when you have n!=2*a, is an even number. Therefore, a must be a odd number although if odd then it is not a perfect square since. There is always a prime factor(b) inside. b has no factors less than its self and one.

For n! to be square, n-(n-1) n-(n-2) ...*(n-1) must equal n. This clearly never happens so no n causes n! to be a perfect square.