# When is n! + 5 a Perfect Cube?

Posted by Jason Polak on 18. January 2016 · Write a comment · Categories: math

I like numbers. And, when I saw that I had just pased $125$ posts, I thought I’d post something really short on $125$, since it’s a pretty cool number. The most obvious property of it is that it’s a perfect cube: $5^3 = 125$. In fact, it’s the only perfect cube of the form $n! + 5$ (I wanted to use an exlamation point at the end of this sentence, but I think it’s best to avoid this kind of punctuation).

Why is this? Suppose $n \geq 10$. Then $n! + 5 = 5(n!/5 + 1)$, and $n!/5$ is an integer and a multiple of $5$. So, $n!/5 + 1$ is not, and hence $n! + 5$ only has one factor of $5$ in it, whereas in a perfect cube, the power of $5$ in its factorisation has to be a multiple of three.

That leaves $n! + 5$ for $n=0,\dots, 9$. So we have to check that none of $6,7,11,29,725,5045,40325, 362885$ are perfect cubes. This is clear for $6,7,11,29$. The other numbers end in five, and hence are a multiple of five, so they would need to be multiples of 125. But multiples of 125 can end only in: 000, 125, 250, 375, or these numbers plus 500: and none of our candidates do! (Oops)

I think it’s pretty strange that proving that $n! + 5$ can only be a perfect cube at most once is pretty straightforward, yet proving that $n!$ is a perfect square only for $n=0,1$ is actually a very tricky proof and seems to require Bertrand’s postulate.