# Can n! be a Square Number?

### 2019-09-11

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.