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.
 A: The statement has nothing to do with sixes and sevens, but just depends on the fact that the exponent $n$ (replacing $6$) is indivisible by the prime $p$.
Your statement has to do rather with whether $p^k\Vert(x-y)$, equivalently $x-y=p^ku$, where $p\nmid u$, equivalently $x=y+p^ku$.
Assuming this for the moment, we raise to the $n$-th power,
\begin{align}
x^n&=y^n+np^kuy^{n-1}+p^{2k}z\\
&=y^n+p^k\bigl(nuy^{n-1}+p^kz\bigr)\\
x^n-y^n&=p^k\bigl(nuy^{n-1}+p^kz\bigr)\,,
\end{align}
where $p^{2k}z$ collects all the other terms from the binomial expansion. Now $nuy^{n-1}$ is indivisible by $p$, and you see that in fact the totality of what’s between the parentheses is also indivisible by $p$. This shows one direction of your implication. But you also see that as long as $p\mid(x-y)$, the full divisibility by $p^k$ of $x-y$ and of $x^n-y^n$ are the same. This proves both directions of your implication.
Those more deeply experienced in number theory will recognize that the proof above is $p$-adic in inspiration.
A: 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.
A: $(\implies)$
Let $p^k \mid \mid (x-y)$. Then $p^k \mid (x-y)$ with $k$ is the greatest positive integer such that $x-y = p^km$ for some nonzero integer $m$. Notice that we must have $p \nmid m$, because otherwise we would have $m=pc$ for some nonzero integer $c$ such that $x-y=p^{k+1}c$, which implies $p^{k+1} \mid (x-y)$, contradicting the maximality of $k$.
For the convenience, let's write $A=x^5+\cdots+y^5$.
Now, notice that
$$x^6-y^6 = (x-y)A= p^kmA = p^k(mA).$$
Hence, $p^k \mid (x^6-y^6)$. The goal is to show that $p^k \mid \mid (x^6-y^6)$. In other words, we want to show that $p^{k+1} \nmid (x^6-y^6)$.
If $p \mid A$, then $A=ps$ for some nonzero integer $s$. Hence,
$$x^6-y^6=(x-y)A=p^km(ps)=p^{k+1}ms,$$ which gives $p^{k+1} \mid (x^6-y^6)$, which is not being our goal. So, $p \nmid A$. It follows that $p^{k+1} \nmid (x^6-y^6)$. To this end, suppose for contradiction that $p^{k+1} \mid (x^6-y^6)$. Then $$x^6-y^6 = p^{k+1}s = p^kps = (x-y)A = p^kmA,$$ for some nonzero integer $s$. Hence, $p^kps=p^kmA$, which implies $ps=mA$. It follows that $p \mid m$ or $p \mid A$. Since $p \nmid m$, then we must have $p \mid A$, contradicting the assumption that $p \nmid A$. Thus,
$$p^{k+1} \nmid (x^6-y^6).$$
This means that $p^k \mid \mid (x^6-y^6)$, and so we're done. $\Box$
$(\impliedby)$
Let $p^k \mid \mid (x^6-y^6)$, i.e.,
$k$ is the greatest positive integer such that
$p^{k} \mid (x^6-y^6)$. We want to show that $p^k \mid \mid (x-y)$, i.e., $p^{k+1} \nmid (x-y)$.
First way: Suppose for contradiction that $p^{k+1} \mid (x-y)$. Then $x-y=p^{k+1}s$ for some nonzero integer $s$.
Now, let $A=x^5+\cdots+y^5$.
Hence,
$$x^6-y^6 = (x-y)A = p^{k+1}sA.$$
Clearly, $A$ is an integer. Thus, $sA$ also being an integer. Therefore, $p^{k+1} \mid (x^6-y^6)$, a contradiction to our assumption that $p^{k+1} \nmid (x^6-y^6)$. Hence, $p^{k+1} \nmid (x-y)$. Thus, $p^k \mid \mid (x-y). \qquad \Box$
Second way: Let $p^k \mid \mid (x^6-y^6)$. Then $x^6-y^6=p^km$ for some nonzero integer $m$. Notice that we must have $p \nmid m$, because otherwise we would have $m=pj$ such that $x^6-y^6=p^{k+1}j$ for some nonzero integer $j$, contradicting the maximality of $k$. Now, let $A=x^5+\cdots+y^5$. We have
$$x^6-y^6 = p^km = (x-y)A.$$
Hence $p^k \mid (x-y)$ or $p^k \mid A$.
If $p^k \mid A$, then $A=p^ks$ for some nonzero integer $s$. Since $p \mid (x-y)$, then $x-y=pt$ for some nonzero integer $t$. Hence
$$x^6-y^6=(x-y)A=pt(p^ks)=p^{k+1}st.$$
Thus, $p^{k+1} \mid x^6-y^6$, contradicting our assumption that $p^k \mid \mid (x^6-y^6)$. Hence, $p^k \nmid A$. Therefore, $p^k \mid (x-y)$. If $p^{k+1} \mid (x-y)$, then $x-y=p^{k+1}r$ for some nonzero integer $r$. Hence
$$x^6-y^6=p^km=p^{k+1}rA,$$ which gives $p \mid m$, which contradicts the maximality of $k$. Thus, $p^{k+1} \nmid (x-y)$. Therefore, $p^k \mid \mid (x-y)$, and we are done. $\Box$
