# Can n! be a Square Number?

### 2019-09-11

I read about this simple yet interesting question the other day:

Can

`$n!$`

be a square number?

Of course, here we implcitly assume that `$n$`

is a positive integer. Moreover, we’re aware of the trivial case when `$n=1$`

(and yes it holds in that case). The problem seems easy at first glance but turns out not at all. There are a bunch of ways for you to factorize `$n!$`

into prime numbers, to construct a pattern of products in the form of a square number, etc. Some may work, some not.

# Original Problem

The best solution is given by the Bertrand’s postulate, which is a theorem that states for any positive integer `$n>1$`

we’ll always have at least one prime number `$p$`

s.t. `$n<p<2n$`

. The proof is not easy and the statement is just as strong — With Bertrand’s postulate we know there’s always a prime `$p$`

that satifies `$\lfloor n / 2\rfloor < p < n$`

and thus in no way can we find another number `$m<n$`

that has factor `$p$`

. Being a prime number, `$p$`

itself is not a square either, so we conclude `$n!$`

can never be a square number for any `$n>1$`

.

# Generalized Problem

Now that we’ve successfully arrived at a conclusion for `$n!$`

, let’s try another problem that highly resembles the one we just solved:

Can product of

`$n$`

consecutive integers be a square number?

Again we’re here neglecting the trivial cases when `$n=1$`

‘cause that then is a problem of whether the only number is a square or not. When `$n>1$`

, the resemblance is apparent: when we consider the first `$n$`

positive integers, the problem falls back to the first one. However, how are we gonna answer this question in general? An educated guess would be a “no” with the conclusion from the first problem and the straightforward observation for `$n=2$`

. When `$n=2$`

, we’re basically asking if `$k\cdot(k+1)$`

can be a perfect square for positive `$k$`

. It can’t, because `$k^2<k\cdot(k+1)<(k+1)^2$`

and there’s simply no square number in between of two consecutive squares.

When `$n=3$`

, it’s a little trickier but still we can handle it. Notice by Euclidean, `$k$`

is coprime with both `$k-1$`

and `$k+1$`

respectively and therefore also with `$k^1-1$`

. This means `$k$`

shares no common factors with `$k^2-1$`

. Hence the product of the two has no duplicate factors and cannot be a perfect square.

When `$n=4$`

, let `$x=k+1/2$`

and we know the product can be written as

$$ \begin{align} (n-1)\cdot n\cdot (n+1)\cdot (n+2) &= \left(k-\frac{3}{2}\right)\left(k-\frac{1}{2}\right)\left(k+\frac{1}{2}\right)\left(k+\frac{3}{2}\right)\\&= \left(k^2-\frac{9}{4}\right)\left(k^2 - \frac{1}{4}\right) \\&= \left(k^2-\frac{5}{4} - 1\right)\left(k^2 - \frac{5}{4} + 1\right) = \left(k^2- \frac{5}{4}\right)^{!!2} - 1 \end{align} $$

and thus by Euclidean (again) this product cannot be a perfect square. What about `$n=5$`

? What about larger and even general `$n>1$`

? It turns out to be an extremely hard one to answer but luckily, Paul Erdös and John Selfridge gave a rigorous proof in their paper published back in 1975. The conclusion, namely that the product of any two or more consecutive positive integers is never a perfect square, is merely a special case of their first result. Actually, theorem 1 in the paper states that the product of two or more consecutive positive integers is never a **power**, i.e. can never be represented in the form of `$p^t$`

where both `$p$`

and `$t$`

are positive integers.