# Numerical oddities…

If, like me, you consider yourself a “serious” ( ™ ) arithmetician, you may also have started spurning at an early age the type of numerical coincidences, sometimes found in dictionaries of remarkable numbers, that state that, for instance, $n=5$ is the only integral solution to the equation
$n^{n-2}=n!+n.$

However, maybe I shouldn’t be so dismissive. Today, after a discussion with P-O Dehaye, which was certainly serious enough since the Ramanujan $\tau$ function featured prominently, the question came to understand the solutions in $\mathbf{Z}/5\mathbf{Z}$ of the system
$v_0+v_1+v_2+v_3+v_4=0,$
$v_0^2+v_1^2+v_2^2+v_3^2+v_4^2=0,$
(everything modulo $5$).

This sounds rather innocuous, but something I find rather amusing happens: you can check for yourself that the solutions are either (i) diagonal (all $v_i$ are equal) or (ii) a permutation of $\mathbf{Z}/5\mathbf{Z}$ (i.e., no two $v_i$ coincide). Why is that? Well, first both (i) and (ii) do describe solutions. Clearly (( ™ ) again) there are $5^{5-2}=125$ solutions in total; there are $5$ diagonal ones, and $5!=120$ permutations; since
$5^3=5!+5,$
there can be no other solution.

Is there a more direct proof?

## Published by

### Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

## 3 thoughts on “Numerical oddities…”

1. Given a solution in which two of the $v_i$ are equal, permute it so that they are the first two terms and translate it so that they are both zero (the solutions are invariant under the translation $v_i \to v_i + 1$). Then $v_2^2 + v_3^2 + v_4^2 = 0$, but this is easily seen to be impossible.

(The group of solutions has a huge group of symmetries. There are commuting actions of $F_{20}$ and $S_5$, that I can see.)

2. Nice! (Though of course it’s not just the equation v_2^2+v_3^2+v_4^2=0 that has no non-trivial solutions — it does, e.g., (0,1,2) –, it is the system of this equation with v_2+v_3+v_4=0).

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