Lifting the Exponent Lemma

I am referring to page 3 of the linked paper. In the paper, the author proves that $v_p(x^p-y^p) = v_p(x-y) +1$. To prove this, he first proves that p divides $\frac{x^p-y^p}{x-p}$. The proof for this makes sense to me although the reason behind his next step is quite unclear to me. For some reason, he proves that $p^2$ does not divide $\frac{x^p-y^p}{x-p}$. What is the purpose of doing this? Isn't it obvious that $p^2$ can't divide $px^{p-1}$ if $p$ can't divide $x$?


The reason why he proves that $p^2$ doesn't divide $\frac{x^p-y^p}{x-y}$ is to prove that $p$ is the highest power of $p$ dividing the integer.

I feel that the confusion stems from one of the first line in the proof where we have that $x^{p-1}y + \cdots xy^{p-1} \equiv px^{p-1} \pmod p$. Note that here we have $p$ as a modulo and not $p^2$. Thus this isn't enough to conclude that the number on the left is equal to $px^{p-1}$ modulo $p^2$. Therefore we need the extra work.

  • $\begingroup$ aaaah i see! Thanks! $\endgroup$ – Dude156 Jul 31 '18 at 1:59

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