# Proving largest power of $(a^2-ab+b^2)$ that divides $a^p-b^p-(a-b)^p$ for odd prime $n$

It is obvious that for odd $$n \in \Bbb N$$, $$a^n-b^n-(a-b)^n$$ is divisible by $$ab(a-b)$$ (with $$n=1$$ being a special case in which $$a^n-b^n-(a-b)^n$$ is zero). This can be viewed as a fact about integers, but is more strongly a fact about these polynomials.

It is apparently also the case that for prime $$p > 3$$, it is also divisible by $$p$$ and there is another simple factor $$\left. (a^2-ab+b^2) \right| a^p-b^p-(a-b)^p$$ and in particular, if $$p = 6k-1$$ for some $$k\in\Bbb N$$ then $$a^p-b^p-(a-b)^p$$ is divisible by $$(a^2-ab+b^2)$$ but not by $$(a^2-ab+b^2)^2$$; while if $$p = 6k+1$$ then $$a^p-b^p-(a-b)^p$$ is divisible by $$(a^2-ab+b^2)^2$$.

This property does not depend on $$p$$ being prime: The remaining part of the statement is that if odd $$n$$ is divisible by $$3$$ then $$a^n-b^n-(a-b)^n$$ is not divisible by $$(a^2-ab+b^2)$$.

So I want to prove that for all $$k\in \Bbb N$$

$$a^{6k+1}-b^{6k+1}-(a-b)^{6k+1} = (6k+1) a b (a-b) (a^2-ab+b^2)^2 P_1(a,b)$$ for some polynomial $$P_1(a,b) \in \Bbb Z[a,b]$$, and $$a^{6k+5}-b^{6k+5}-(a-b)^{6k+5} = (6k+5) a b (a-b) (a^2-ab+b^2) P_5(a,b)$$ for some polynomial $$P_5(a,b)\in \Bbb Z[a,b]$$ which itself is not a multiple of $$(a^2-ab+b^2)$$, and $$a^{6k+3}-b^{6k+3}-(a-b)^{6k+3} = (6k+3) a b (a-b) P_3(a,b)$$ where $$P_3(a,b)$$ is not divisible by $$(a^2-ab+b^2)$$.

In trying to prove this, I started by noting that $$(a^2-ab+b^2) = (a-\omega b) (a - \omega^2b)$$ where $$\omega$$ is a non-trivial cube root of unity. I was hoping that this would shed some light on why multiples of $$3$$ would not have the divisibility, but I can't see how this follows.

Is there some more powerful technique that will make the indicated statements "fall out" or at least be easier to prove?

• All relevant polynomials are homogeneous; you can use that to reduce by one variable. – user44191 Apr 23 '19 at 19:27
• Yes, and that reduces the problem to proving some rather messy relations among sums of binomial coefficients. This might be how to do it, because powerful techniques like those of Gosper/Zeilberger might be brought to bear. But the relations are more than a bit hairy. – Mark Fischler Apr 23 '19 at 19:50
• If you reduce the problem to one variable, as suggested by user44191, then the question becomes whether $\omega$ is a root resp. a multiple root. I guess that $\omega$ should be a 6th root of unity, so $1-\omega=\omega^5$ , so all occurring powers can be computed easily. – Jan-Christoph Schlage-Puchta Apr 23 '19 at 20:42
• You will probably be interested to read Lemma 4.2 and Proposition 4.3 of my paper with Boyd and Thom. The methods there show, for example, that the $6k+1$ divisibility by $(a^2-ab+b^2)^2$ is essentially due to the fact that $x^{6k+1}-(x-1)^{6k+1}-1$ has a cyclotomic factor $x^2-x+1$, and $a^2-ab+b^2$ is the resultant of that factor with its cofactor. – Greg Martin Apr 24 '19 at 2:07

$$\omega$$ in what follows is the sixth root of unity.

As user44191 suggested, lets reduce this question to the 1-variable. For the first statement it is enough to prove that

$$x^{6k+1} - 1 - (x-1)^{6k+1}$$

has a root $$\omega$$ of order 2. It is clear that $$\omega^6 = (\omega-1)^6 = 1$$ (because $$\omega-1$$ is the third root of unity. From this follows

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

The same proof works for $$6k+5$$, and also closes $$6k+3$$ case (shows that $$\omega$$ is not a root in this case)

Now we need to check what happens with the derivative of this polynomial.

For 6k+1 case:

$$(6k+1) \omega^{6k} - (6k+1)(\omega - 1)^{6k} = -1$$. Root (hence, $$\omega$$ is a root of order $$2$$)

For 6k+5 case:

$$(6k+5) \omega^{6k+4} - 1 - (6k+5)(\omega - 1)^{6k+4} = (6k+5)(-\omega) - (6k+5)(\omega-1) = 6k+5$$. Not a root.

• To complete the analysis you must also check that in the $6k+5$ the second derivative is not zero so $\omega$ is not a root of order $3$ or more. – Noam D. Elkies Apr 23 '19 at 21:35
• Your last statement "Not a root" is a bit misleading, in that it is saying that the derivative is not zero, not that $\omega$ is "not a root." Also, in the derivative calculation, did you mean $= -1$ on the RHS or $=0$? But your answer gets the proof done nicely. I did not think to work with sixth roots rather than cube roots of unity. – Mark Fischler Apr 23 '19 at 21:40