Prove that $(a-b)^n\mid (a^n-b^n) \iff n=1$ under given conditions Suppose that $a,b,(a-b)$ are pairwise co-prime (i.e. $a\perp b\perp (a-b)\perp a$), and that $\frac{a}{2}<b<a$, where $a$ and $b$ are both positive integers greater than $2$.  Let $n$ be odd.  Prove or disprove that $(a-b)^n\mid (a^n-b^n)$ iff $n=1$.  
I can see that one possible implication is that
$$a\equiv b   \pmod{(a-b)^n},$$
from which $(a-b)^n\mid (a-b)$ which is true iff $n=1$.
But is this the only implication?
Note $a\perp b$ means that $\rm{gcd}(a,b)=1$.
 A: I guess OP missed the obvious counterexample $a-b=1$. So, I will assume $a-b\ge 2$, in particular, it has a prime divisor.
The case $n=1$ is obvious. Suppose $n\ge3$. Let $p$ be an odd prime dividing $a-b$, and let $v_p(m)$ denote the power of $p$ in $m$. Then, by Lifting the Exponent lemma
$$v_p(a^n-b^n)=v_p(a-b)+v_p(n)$$
However, $(a-b)^n|a^n-b^n$ implies
$$n\cdot v_p(a-b)\le v_p(a^n-b^n)=v_p(a-b)+v_p(n)$$
So, we have
$$(n-1)v_p(a-b)\le v_p(n)\implies p^{n-1}\le p^{(n-1)v_p(a-b)}\le p^{v_p(n)}\le n$$
However, $p\ge 3,n\ge 3$ gives contradiction. So, $a-b$ has no prime odd prime divisor. So, $a-b=2^m$ for some $m$.
As expected, we will continue with LTE's special case for $p=2$. Let $k=v_2(n)$, then, in particular, $2^k\le n$. If $m\ge 2$, then, $4|a-b$, so,
$$v_2(a^n-b^n)=v_2(a-b)+v_2(n)=m+k$$
Moreover, $(a-b)^n|a^n-b^n$ gives
$$mn=n\cdot v_2(a-b)\le v_2(a^n-b^n)=m+k\implies 2(n-1)\le m(n-1)\le k$$
$$\implies 4^{n-1}\le 2^k\le n$$
As $n\ge 3$, we still have contradiction.
The last case is $a-b=2$, then, $a,b$ are both odd, so, they have different residues modulo $4$. As $n$ is odd,
$$a^n\equiv a\not\equiv b\equiv b^n\pmod4$$
So, $4\nmid a^n-b^n$. Thus, $(a-b)^n=2^n\nmid a^n-b^n$.
