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I'm working on Fermat property that says that :

$\forall \; n\geq3, \; \nexists \;(a;b;c)$ such as $ a^n+b^n=c^n $

I am asked to prove that it is necessary to use congruences greater or equal to 7 in order to prove that :

$4693^{20}+4110^{20}\neq4709^{20}$

I checked that for modulo 7, the equality doesn't hold but I don't know how to prove it in the general case.

Thank you

Romain (from France)

philok
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    The title is not correct. It is not true that $a^n+b^n=c^n$ for all $n<3$. – Dietrich Burde Nov 08 '20 at 13:19
  • Thank you, did I correct it right ? – philok Nov 08 '20 at 13:24
  • How to prove what in the general case? I think, but I am not at all sure, that the point is for you to show that congruences mod $2,3,5$ don't suffice. It's phrased poorly. Obviously you don't need congruences at all...if you wanted to, you could just write out both sides and check. – lulu Nov 08 '20 at 14:04
  • You are right, I need to show that congruences below 7 don't suffice. Do I have to check for 4 and 6 also since two numbers can be equal modulo 2 but not modulo 4 ? – philok Nov 08 '20 at 14:22
  • And I was also wondering if there was a way to directly say that we have to look at modulos above 7 in order to demonstrate the inequality, without having to test with 2,3,4,5,6. – philok Nov 08 '20 at 14:29
  • Yes, you should check $4$ as well. No need to check $6$ since a solution mod $2$ and mod $3$ would give you a solution mod $6$. I don't think there's a short cut, but none of the computations are difficult. – lulu Nov 08 '20 at 14:32
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    Should say, your title doesn't make any sense. More broadly, there is no point referring to Fermat's Last Theorem at all. You have a particular equation that you need to test, that's it. In theory you could invoke Fermat to settle it, but that would be preposterous overkill. – lulu Nov 08 '20 at 14:33

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