# false counterexamples to Fermat's last theorem

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Type of Math Object:
Example
Major Section:
Reference

## Mathematics Subject Classification

### Another TeX verbatim glitch

I don't know what's wrong with this entry. My top guess is: yet another glitch with verbatim. I'm giving a couple of friends access if they want to take a try at fixing it.

### Re: Another TeX verbatim glitch

BTW, the invalidity of the second "equation"
6107^6+8919^6 = 9066^6
is seen by inferring the last digit of both sides:
LHS is modulo 10 congruent with 7^6+9^6 = 9^3+1^3 = 9+1 = 0,
RHS modulo 10 with 6^6 = 6^3 = 6.
Regards,
Jussi

### Re: Another TeX verbatim glitch

These Fermat "counter examples" demonstrate the power of elementary modular algebra. The first "equation" fails modulo 2, as noted by the author himself. The second one fails module 10, as noted by Pahio. The third one fails modulo 9: both LHS numbers are divisible by 9, the RHS is not.

### Re: Mod arithmetic disproves Fermat counterexamples

> These Fermat "counter examples" demonstrate the power of elementary
> modular algebra. The first "equation" fails modulo 2, as noted by the > author himself. The second one fails module 10, as noted by Pahio. The > third one fails modulo 9: both LHS numbers are divisible by 9, the RHS > is not.
>

"As noted by the author HERself."

Overall, I agree. In fact, this is an opportunity to link to the PM entry that proves that if $a \equiv b \mod c$ then $a^n \equiv b^n \mod c$ for $n > 0$. Only... I can't find that entry... is it missing?

### Re: Mod arithmetic disproves Fermat counterexamples

Yes, but it wouldn't hurt to say that exponentiation is one such function with integer coefficients. In fact, because exponentiation is not typically expressed in function notation, it is all the more pressing to specifically mention it.