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Assume $x,y\in\mathbb{N}$ and $p$ - prime, $p\ge 5$.

$5xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^5-x^5-y^5$ (in fact, there is equality).

$7xy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^7-x^7-y^7$ (there is equality also).

$11xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^{11}-x^{11}-y^{11}$ but $11xy(x+y)(x^2+xy+y^2)^2$ does not.

It is easy to prove that $pxy(x+y)$ divides $(x+y)^p-x^p-y^p$.

Does $pxy(x+y)(x^2+xy+y^2)$ divide $(x+y)^p-x^p-y^p$ always?

Is there a characterisation of $p$'s for which $pxy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^p-x^p-y^p$ ?

Edit:

For $p=17$ there is no divisibility by $(x^2+xy+y^2)^2$, similarily as for $p=11$.

tong_nor
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2 Answers2

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By dehomogenizing the polynomials, it suffices to work with $f_p (x) = (x+1)^{p} - x^p - 1$. Let $g(x) = x^2 + x + 1$ and for any polynomial $f$, let $\text{ord}_g (f)$ denote the maximum power of $g$ that divides $f$. We claim $\text{ord}_g (f_p) = 1$ if $p \equiv 5 \pmod{6}$ and $\text{ord}_g (f_p) = 2$ if $p\equiv 1\pmod{6}$.

Let $\omega = (-1+\sqrt{-3})/2$ be the root of $g(x)$ with negative real part. We know $\omega^3 = 1$ and $\omega+1$ is a root of $x^2 - x + 1$, so $(\omega+1)^3 = -1$.

If $p \equiv 5 \pmod{6}$, writing $p=6k+5$ we find $$f_p (\omega) = (\omega + 1)^{6k+5} - \omega^{6k+5} - 1 = -(\omega+1)^2 - \omega^2 - 1 = -2(\omega^2 + \omega + 1) = 0.$$ Thus, $\omega$ is a root of $f_p$ and it follows $\overline{\omega}$ also is, hence $g$ divides $f_p$. But $\omega$ is not a root of $f_p'(x) = p(x+1)^{p-1} - px^{p-1}$ since $$f_p'(\omega) = p(\omega+1)^{6k+4} - \omega^{6k+4}) = p(-(\omega+1) - \omega) = -p(2\omega+1) \neq 0.$$ Thus, $g$ does not divide $f_p'$ and it follows $\text{ord}_g (f_p) = 1$.

However, if $p\equiv 1 \pmod{6}$, writing $p=6k+1$ we find $$f_p (\omega) = (\omega+1)^{6k+1} - \omega^{6k+1} - 1 = (\omega+1) - \omega - 1 = 0$$ and $$f_p'(\omega) = p((\omega+1)^{6k} - \omega^{6k}) = 0,$$ hence $g^2$ divides $f_p$. But $f_p ''(x) = p(p-1)(x+1)^{p-2} - p(p-1)x^{p-2}$ and $$f_p ''(\omega) = p(p-1)((\omega+1)^{6k-1} - \omega^{6k-1}) = p(p-1)\left(\frac{1}{\omega+1} - \frac{1}{\omega} \right) = p(p-1) \frac{-1}{\omega^2 + \omega} = p(p-1) \neq 0.$$ So $g^3$ does not divide $f_p$ and we conclude $\text{ord}_g (f_p) = 2$.

Sameer Kailasa
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Solving for $x$ in $x^2+xy+y^2=0$, we obtain $$x = \frac{-y \pm \sqrt{y^2-4y^2}}{2}=\frac{-1 \pm i\sqrt{3}}{2}y$$ Call one solution (the one with positive imaginary part) $\xi y$, the other $\xi'y$, so that $\xi^3 = \xi'^3 = 1$. Then it's easy to show that $x = \xi y$ and $x = \xi'y$ are roots of $f(x,y)=(x+y)^p-x^p-y^p$, and so $(x-\xi y)(x-\xi'y)=x^2+xy+y^2$ divides $f$.

$(x^2+xy+y^2)^2$ will divide $f$ precisely when $x = \xi y, \xi' y$ are multiple roots of $f$, i.e. when they are roots of $\frac{\partial f}{\partial x} = p((x+y)^{p-1}-x^{p-1})$. Letting $x=\xi y$, we have $$0 = (\xi y + y)^{p-1}-(\xi y)^{p-1}$$ $$0 = (1+\xi)^{p-1}-\xi^{p-1}$$ $$\xi^{p-1}=(1+\xi)^{p-1}$$ $$\xi = \zeta_{p-1}(1+\xi)$$ ($\zeta_{p-1}$ is some $(p-1)^{th}$ root of unity, not necessarily primitive) $$\zeta_{p-1} = \frac{\xi}{1+\xi} = \frac{1}{2}+\frac{\sqrt{3}}{2}i$$ That last number is a complex cube root of $-1$, and a primitive cube root of $6$. Had we used $\xi'$ here, we would have gotten the other complex cube root of $-1$. In order for $\zeta_{p-1}$ to be a primitive sixth roots of $1$, we need $p-1$ to be a multiple of six, and so $p \equiv 1$(mod $6$). This is also sufficient.

florence
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