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.

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.
