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I was just playing with the factorial and the modulo function. I just observed this interesting property. I was using a calculator

$$13!\equiv 13\times 12\pmod{169}\\ 17!\equiv 17\times 16\pmod{289}$$

It is easily verifiable that this works for $2,3,5,7,11$ also.

I conjecture that for any prime $p$, $$p!\equiv (p)\times (p-1)\pmod{p^2}$$

How does one go about proving it? and by the way is this well known or anything?

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By Wilson's Theorem we have $(p-1)!\equiv -1\equiv p-1\pmod{p}$. Your conjectured result is obtained by multiplying through by $p$.

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  • $\begingroup$ How do you prove that the last operation is valid.(The one where you multiply everything by $p$). $\endgroup$ – The Cryptic Cat Apr 3 '16 at 3:30
  • $\begingroup$ Oh! It follows from the definition of the congruent modulo function. +1 $\endgroup$ – The Cryptic Cat Apr 3 '16 at 3:34
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    $\begingroup$ Yes, it is very close to Wilson's Theorem. We have $p(p-1)!-p(p-1)$ is divisible by $p^2$ if and only if $(p-1)!-(p-1)$ is divisible by $p$. $\endgroup$ – André Nicolas Apr 3 '16 at 3:42
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    $\begingroup$ By the way, Leibniz was the first person to state the result that would later be called Wilson's Theorem. It is not known whether he had a proof. For sure Wilson did not! The first known proof is by Lagrange. $\endgroup$ – André Nicolas Apr 3 '16 at 3:48

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