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

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
(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
there can be no other solution.

Is there a more direct proof?

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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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