# Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way?

I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p - b^p=p\cdot(ab)^{\alpha_1}(a+b)^{\alpha_2}(a^2+ab+b^2)^{\alpha_3}P_{\omega}(a,b)$ for any prime number p where $P_{\omega}(a,b)$ is an irreducible homogenous polynomial of degree $\omega$ and $\alpha_1 + \alpha_2 + \alpha_3 + \omega = p-1$. I wonder if it's true in general.

Here are some examples:

$(a+b)^2 - a^2 - b^2 = 2ab$

$(a+b)^3 - a^3 - b^3 = 3ab(a+b)$

$(a+b)^5 - a^5 - b^5 = 5ab(a+b)(a^2+ab+b^2)$

$(a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2+ab+b^2)^2$

$(a+b)^{11}- a^{11} - b^{11} = 11 ab(a + b)(a^2 + a b + b^2)(a^6 + 3 a^5 b + 7 a^4 b^2 + 9 a^3 b^3 + 7 a^2 b^4 + 3 a b^5 + b^6)$

I've proved the following statements:

For any prime number p:

$p|(a+b)^p - a^p-b^p$

$ab|(a+b)^p - a^p-b^p$

For any odd prime number:

$(a+b) | (a+b)^p - a^p-b^p$ because $(a+b)^p - a^p-b^p$ vanishes when $a=-b$ if p is odd.

For any prime number $p \geq 5$:

$(a^2 + ab + b^2) | (a+b)^p - a^p - b^p$

Proof: Any prime number greater than 3 is of the form $p=6k+1$ or $p=6k+5$. On the other hand, if we set $\Large \omega= e^\frac{2\pi i}{3}$ we have: $1 + \omega + \omega^2 = 0$, therefore

$(1+\omega)^3 = 1 + 3\omega + 3\omega^2 + \omega^3 = 2 + 3(\omega+\omega^2) = -1$

$(1+\omega^2)^3 = 1 + 3\omega^2 + 3\omega^4 + \omega^3 = 2 + 3(\omega^2+\omega) = -1$

We have $a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)$ in $\mathbb{Z}[\omega]$, I'll show that both $a-b\omega$ and $a-b\omega^2$ divide $(a+b)^p - a^p - b^p$ for every $p>3$:

if $p=6k+1$ and we replace $a=b\omega$ then:

$(a+b)^p=(b\omega+b)^{6k+1} = b^{6k+1} (1+\omega)^{6k} (1+\omega) = b^{6k+1}(1+\omega)$ $(a+b)^p - a^p - b^p = b^{6k+1}(1+\omega) - (b\omega)^{6k+1} - b^{6k+1} = b^{6k+1}(1+\omega) - b^{6k+1}(\omega+1) = 0$

If $p=6k+5$ and we replace $a=b\omega$ then:

$(a+b)^p=(b\omega+b)^{6k+5} = b^{6k+5} (1+\omega)^{6k} (1+\omega)^5 = (-1)b^{6k+1}(1+\omega)^2$ $(a+b)^p - a^p - b^p = (-1)b^{6k+5}(1+\omega)^2 - (b\omega)^{6k+5} - b^{6k+5} = (-1)b^{6k+5}(1+\omega)^2 - b^{6k+5}(\omega^2+1) = p^{6k+5}(-1-2\omega-\omega^2-\omega^2-1)=0$

The same could be shown for $a=b\omega^2$.

Therefore both $(a-b\omega)$ and $(a-b\omega^2)$ are factors of $(a+b)^p-a^p-b^p$ but I'm not sure if that is sufficient to conclude $a^2+ab+b^2=(a-b\omega)(a-b\omega^2) | (a+b)^p - a^p - b^p$

So, I guess I have proved that $p$, $ab$, $a+b$ and $a^2+ab+b^2$ all divide $(a+b)^p-a^p-b^p$, but I have no idea how to show that $P_{\omega}(a,b)$ must be irreducible. I have verified this conjecture up to $p=97$ and it's true I think. Any ideas on how to prove that?

• Why don't you begin with the binomial theorem? – Avitus Aug 25 '13 at 19:50
• @Avitus: I don't see how that could help. – user66733 Aug 25 '13 at 20:12
• a tiny remark ($p\neq2$): $\alpha_1=\alpha_2=1$, $\alpha_2=2$ for $p\equiv 1$ mod $3$, otherwise $\alpha_2=1$ (follows by looking for multiple roots). But no idea about irreducibility of the rest. – user8268 Aug 25 '13 at 20:17
• @user8268: Yea, I also guessed that pattern about the multiplicity of $a^2+ab+b^2$. Do you have a proof for that? If yes, please write it down here, it's definitely better than nothing.I guess the irreducibility part might be quite hard to prove if it's true. – user66733 Aug 25 '13 at 20:21
• here: ($a=x,b=1$): $f(x):=(1+x)^p-1-x^p$, $f'=p((1+x)^{p-1}-x^{p-1})$. $f'$ has no double roots, as $f'=px^{p-1}(((1+x)/x)^{p-1}-1)$. If $c$ is a double root, $f(c)=f'(c)=0$, equivalently $c^{p-1}=1$ and $(1+c)^{p-1}=1$. At least $c$ is on the circles $|x|=1$ and $|1+x|=1$, their intersection is $\omega,\bar\omega$, so these are the possible multiple roots. But $\omega^{p-1}=(1+\omega)^{p-1}=1$ iff $p\equiv1$ mod $3$. – user8268 Aug 25 '13 at 20:35

It is true in general. These are often called Cauchy polynomials — see, for example, Fermat's Last Theorem for Amateurs, Chapter VII. The power of $a^2+ab+b^2$ is either $1$ or $2$ as $p \equiv -1$ or $+1$ modulo $6$, respectively. There are many proofs in the literature (Cayley, Glaisher, etc.).