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, $latex n=5$ is the only integral solution to the equation
$latex n^{n-2}=n!+n.$
However, maybe I shouldn’t be so dismissive. Today, after a discussion with P-O Dehaye, which was certainly serious enough since the Ramanujan $latex \tau$ function featured prominently, the question came to understand the solutions in $latex \mathbf{Z}/5\mathbf{Z}$ of the system
$latex v_0+v_1+v_2+v_3+v_4=0,$
$latex v_0^2+v_1^2+v_2^2+v_3^2+v_4^2=0,$
(everything modulo $latex 5$).
This sounds rather innocuous, but something I find rather amusing happens: you can check for yourself that the solutions are either (i) diagonal (all $latex v_i$ are equal) or (ii) a permutation of $latex \mathbf{Z}/5\mathbf{Z}$ (i.e., no two $latex v_i$ coincide). Why is that? Well, first both (i) and (ii) do describe solutions. Clearly (( ™ ) again) there are $latex 5^{5-2}=125$ solutions in total; there are $latex 5$ diagonal ones, and $latex 5!=120$ permutations; since
$latex 5^3=5!+5,$
there can be no other solution.
Is there a more direct proof?
