Prove that $x^2+px+p^2$ is a factor $(x+p)^n-x^n-p^n$, if $n$ be odd and not divisible by $3$. Question:
Prove that $x^2+px+p^2$ is a factor of $(x+p)^n-x^n-p^n$, if $n$ is odd and is not divisible by $3$.
My approach:
$$(x+p)^n-x^n-p^n=\sum_{r=0}^n\limits {n\choose r} x^{n-r}p^r-x^n-p^n$$ What can I do after that?
 A: You need to show every root of $x^2 + px + p^2$ is a root of $(x + p)^n - x^n - p^n$. The quadratic formula gives that $x = p\omega$ or $p\bar{\omega}$
where $w = -{1 \over 2} + {\sqrt{3} \over 2}i$ is a complex third root of unity. So what you need to show is that for the values of $n$ in question
that 
$$(p\omega + p)^n - (p\omega)^n - p^n = 0$$
You also need this for $\bar{\omega}$ in place of $\omega$ but that will follow by taking complex conjugates of this equation.
Hence you need to show for your values of $n$ that
$$(\omega + 1)^n = \omega^n + 1$$
I won't finish this off but you can reduce this to checking a small list of $n$.
A: First of all substitute $x$ with $py$
Observe $y^3-1=(y-1)(?)=0$
So, $y=w,w$ is a complex cube root of unity
Now $(x+p)^n-x^n-p^n=p^n((1+y)^n-1-y^n)$
Now $f(n)=(1+w)^n-1-w^n=(-w^2)^n-1-w^n$
which is $=-1-w^n-w^{2n}$ if $n$ is odd
If $3\mid n,f(n)=-1-1^n-1^{2n}=?$
Else
$w^n\ne1,f(n)=-\dfrac{1-(w^3)^n}{1-w^n}=0$
Check for $n=3m+1,3m+2$
A: First of all we gotta focus on the phrase
$n$ is odd and not divisible by $3$ ... And general formats of $n$ that satisfy the above condition are
$$6k+1 ;k=0,1,2,3...$$
and $$6k-1 ;k=1,2,3,4... $$
Now roots of $x^2+px+p^2=0$ are $p\omega$ and $p\omega^2$ ( $\omega$, $\omega^2$ are complex cube roots of unity)
Given, $f(x) = {(x+p)^n}-(x^n)-(p^n)$
Using basic properties of cube roots of unity , it's easy to see that
$$f(pω)= (p^n)[-ω^2-ω-1] =0$$
And $$f(pω^2)= (p^n) [ -ω-ω^2-1]=0$$
[ Put $x= p\omega$ and $p\omega^2$ respectively; take $p^n$ common. Then put $n=6k+1$ and $6k-1$ for $x= p\omega$ case then again for $x= p\omega^2$ case respectively. Use basic properties of $\omega$ and
your simple form is ready ]
Therefore $x^2+px+p^2=0$ is a root of $f(x)$.
