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?

  • $\begingroup$ All relevant polynomials are homogeneous; you can use that to reduce by one variable. $\endgroup$
    – user44191
    Apr 23 '19 at 19:27
  • $\begingroup$ 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. $\endgroup$ Apr 23 '19 at 19:50
  • $\begingroup$ 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. $\endgroup$ Apr 23 '19 at 20:42
  • $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$ Apr 23 '19 at 21:35
  • $\begingroup$ 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. $\endgroup$ Apr 23 '19 at 21:40

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