Wondering about correctness of this:



$ \quad \quad \quad \quad p$ be an odd prime

$ \quad \quad \quad \quad \gcd(x,y,z) = 1$

$ \quad \quad \quad \quad xyz \not \equiv 0 \pmod p$

$ \quad \quad \quad \quad a \equiv y/z \pmod p$

$ \quad \quad \quad \quad x^p = y^p + z^p$

Then $p$ is a Wieferich-prime to base $a$.



$ \quad \quad \quad \quad y/z \equiv a \implies y \equiv az \pmod p$

$ \quad \quad \quad \quad \implies y^p \equiv (az)^p \pmod {p^2}$

$ \quad \quad \quad \quad \implies x^p = y^p + z^p \equiv (az)^p + z^p \equiv (a^p + 1)z^p \pmod {p^2}$


$ \quad \quad \quad \quad x \equiv y + z \equiv az + z \pmod p$

$ \quad \quad \quad \quad x^p \equiv (az + z)^p \equiv (a + 1)^pz^p \pmod {p^2}$


$ \quad \quad \quad \quad x^p \equiv (a^p + 1)z^p \equiv (a + 1)^pz^p \pmod {p^2}$

$ \quad \quad \quad \quad a^p + 1 \equiv a + 1 \implies a^p \equiv a \pmod {p^2}$

So our theorem is correct.

  • 1
    $\begingroup$ Uh ... it is impossible to satisfy $x^p=y^p+z^p$ and $xyz\ne 0$ when $p>2$ (Wiles, 1995), and so in particular when $p$ is an odd prime. So your claim is vacuously true. $\endgroup$ – hmakholm left over Monica Jun 24 '17 at 1:34
  • 3
    $\begingroup$ @HenningMakholm True,of course, but I suspect they're looking for an elementary proof of this claim, though ... $\endgroup$ – Noah Schweber Jun 24 '17 at 1:59

How do you conclude in the last line that $a^p+1\equiv a+1\pmod{p^2}$? You've shown that $$(a^p+1)z^p\equiv x^p\equiv(a+1)^pz^p\pmod{p^2},$$ which implies that $a^p+1\equiv(a+1)^p\pmod{p^2}$ because $p\nmid z$. This does not immediately imply $a^p+1\equiv a+1\pmod{p^2}$, as far as I can tell.

  • $\begingroup$ they made a mistake, thanks $\endgroup$ – user451710 Jun 25 '17 at 11:03

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