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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.