Factorial which is Sum of Two Squares

Theorem

The only factorial which can be expressed as the sum of two squares is:

$\ds 6!$

$=$

$\ds 12^2 + 24^2$

We show that for $n \ge 7$, $n!$ cannot be expressed as the sum of two squares.

There exists a prime of the form $4 k + 3$ strictly between $m$ and $2 m$ whenever $m \ge 4$.

Let $n \ge 7$. Then $\ceiling {\dfrac n 2} \ge 4$.

Using the result above, there is a prime $p$ of the form $4 k + 3$ such that:

$\ceiling {\dfrac n 2} < p < 2 \ceiling {\dfrac n 2}$

We then have, by multiplying the inequality by $2$:

$2 \ceiling {\dfrac n 2} < 2 p < 4 \ceiling {\dfrac n 2}$

This gives:

$p < 2 \ceiling {\dfrac n 2} < 2 p$

Which implies:

$p \le n < 2 p$

$n!$ can be expressed as the sum of two squares if and only if each of its prime divisors of the form $4 k + 3$ (if any) occur to an even power.

The inequality above shows that there are no multiples of $p$ which are not greater than $n$ except $p$ itself.

Hence $p$ occurs to an odd power, $1$, in $n!$.

This shows that for $n \ge 7$, $n!$ cannot be expressed as the sum of two squares.

Checking the rest of the factorials we see that the only ones satisfying the criteria are:

$\ds 0! = 1!$

$=$

$\ds 0^2 + 1^2$

$\ds 2!$

$=$

$\ds 1^2 + 1^2$

$\ds 6!$

$=$

$\ds 12^2 + 24^2$

Hence the result.

$\blacksquare$