prove that $\sum_{i+j+k=p}{p\choose i+j+k}x^iy^jz^k\equiv x^p+y^p+z^p\pmod{(x+y)(y+z)(z+x)}$

Question

Let $x,y,z$ be integers and $p$ an odd integer.How can I prove that $$\sum_{i+j+k=p}{p\choose i+j+k}x^iy^jz^k\equiv x^p+y^p+z^p\pmod{(x+y)(y+z)(z+x)}$$

Answer 1

We have to prove that the polynomial $f=(x+y)(x+z)(y+z)$ divides the polynomial $g=(x+y+z)^p-(x^p+y^p+z^p).$ It is enouth to show that all roots of $f$ are roots and for $g.$ Take, for example, a root $x=-y$ and put into $g$. We get $z^p-( (-y)^p+y^p+z^p)=0$ for odd $p$. Thus $g=(x+y)g_1$ for some polynomial $g_1.$ In a similar way working with $x=-z$ and $y=-z$ we get that $g=0 \mod (x+y)(x+z)(y+z).$