<s>[INST] Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$. Definition. Let $n>1$ be an integer and $p$ be a prime. We say that $p^k$ fully divides $n$ and write $p^k \mid \mid n$ if $k$ is the greatest positive integer such that $p^k \mid n$.
Let $p>6$ be a prime. Let $x$ and $y$ be two distinct integers such that $p\nmid x, p\nmid y,$ and $p\mid (x-y)$. Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$.
Attempt:
$(\implies)$ Let $p^k \mid \mid (x-y)$. Then $k$ is the greatest positive integer such that $p^k \mid (x-y)$. Write $x-y=p^km$ for some $m \in \Bbb Z$. Notice that
$$x^6-y^6 = (x-y)(x^5+x^4y+\cdots+y^5) = p^km(x^5+x^4y+\cdots+y^5)=p^k(m(x^5+x^4y+\cdots+y^5)).$$
Hence, $p^k \mid (x^6-y^6)$ and then $p^k \mid \mid (x^6-y^6)$.
Does this approach correct? If not, how to approach it? And how to approach the reverse direction?
Any ideas? Thanks in advanced.
 [/INST] Your attempt so far looks good (as far as I can tell). What you really need to show is that $p\nmid (x^5+\dots+y^5)$ because by now you have only shown that $p^k \mid (x^6-y^6)$. Let's focus on this point, then the reverse direction gets easy. We have $p|(x-y)$ thus $x-y \equiv 0 \mod p$. From this we conclude $x\equiv y \mod p$, so $x^5 \equiv x^4y\equiv x^3y^2\equiv x^2y^3 \equiv xy^4 \equiv y^5 \mod p$. Now $(x^5+\dots + y^5) \equiv 6x^5 \mod p$ and since $p\nmid x$ we have $p\nmid x^5$ and we can conclude $ 6x^5\not \equiv 0 \mod p$. This completes your argument, for the reverse direction consider: $p^k\mid\mid  (x^6-y^6)$ so $(x^6-y^6)=p^km=(x-y)(x^5+\dots +y^5)$. Now since $p\nmid (x^5+\dots y^5)$ we have $p^k \mid (x-y)$ and of course $p^k\mid\mid (x-y)$ because otherwhise we would have $p\mid m$ which is not possible.
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