Divisibility property of $(a+b)^n-a^n-b^n$ Let $n$ be a natural number of the form $n=6k+1$ (while $k$ is a positive integer).
Show that $(a^2+ab+b^2)^2$ divides $(a+b)^n-a^n-b^n$ for all integer numbers $a,b$ (such that $a^2+ab+b^2\ne0$).  
 A: if $\omega$ is a complex cube root of unity, then both $\omega$ and $1+\omega$ are sixth roots of unity. hence for $k \ge 1$$\omega$ is a root of:
$$
(x+1)^{6k+1} - x^{6k+1} - 1 = 0 \tag{1}
$$
likewise for $\omega^2$
but these are exactly the roots of
$$
x^2+x+1=0
$$
this shows that $a^2+ab+b^2$ is a factor of $(a+b)^{6k+1}-a^{6k+1}-b^{6k+1}$ 
however the derivative of the LHS of (1) is $(6k+1)\left((x+1)^{6k}-x^{6k}\right)$
which is also satisfied by $\omega, \omega^2$, so each is, in fact, a double root of the equation (1). hence the result
A: Note: This proof replaces an earlier less elementary version that used the unique factorization property of $\mathbf{Z}[x,y]$.
Let $c = a^2 + ab + b^2$. Since $(a-b)(a^2 + ab + b^2) = a^3 - b^3$, we have $a^3 \equiv b^3$ modulo $c$. It follows therefore that for any $k$ we have $a^{6k} \equiv b^{6k}$ modulo $c$, hence that $c$ divides $a^{6k} - b^{6k}$.
The fact to be proved is trivial for $k=0$. Assume by induction that $(a+b)^{6k+1} \equiv a^{6k+1} + b^{6k+1}$ modulo $c^2$. Calculating modulo $c^2$, we have
$$
\begin{align}
(a+b)^{6k+7} &= (a+b)^6(a+b)^{6k+1} \\
&\equiv \left[ (a+b)^6 - 7abc^2 \right](a^{6k+1} + b^{6k+1})\\
&= \frac{a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) \\
&\equiv \frac {a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) + ab(a-b)(a^2 - ab +b^2)c(a^{6k}-b^{6k}) \\
&= \frac {a^7 + b^7}{a+b}(a^{6k+1} + b^{6k+1}) + ab\frac{a^6-b^6}{a+b}(a^{6k}-b^{6k}) \\
&=a^{6k+7} + b^{6k+7},
\end{align}
$$
where for the first equivalence we've used the induction hypothesis, and for the second the divisibility fact mentioned above. (The desired property is easy to prove separately when  $a+b = 0$. Alternatively, in order for the case $a+b = 0$ to be included above, replace the fractions with the polynomials in $a$ and $b$ to which they're equal.) This completes the proof by induction on $k$. 
