Mathematical Reflections 2013, Issue 4 - Problem J274

Problem:
Let $p$ be a prime and let $k$ be a nonnegative integer. Find all positive integer solutions $(x, y, z)$ to the
equation $$x^k(y-z)+y^k(z-x)+z^k(x-y)=p.$$

Proposed by Alessandro Ventullo.

Solution:

It's easy to see that if $k=0,1$, the equation has no solution. Suppose that $k \geq 2$ and put $f(x)=x^k(y-z)+y^k(z-x)+z^k(x-y)$. Since $f(y)=0$, then $(x-y)|f(x)$ and since the expression is cyclic, we obtain that also $(y-z)$ and $(z-x)$ divide $f(x)$, so $$x^k(y-z)+y^k(z-x)+z^k(x-y)=(x-y)(y-z)(z-x)h(x,y,z), \qquad h \in \mathbb{Z}[x,y,z].$$
Put $x-y=a,y-z=b,z-x=c$. Let $p>2$. Then $a,b,c \in \{\pm 1, \pm p\}$ and $a+b+c=0$, but this is impossibile since $a,b,c$ are all odd. Now, let $p=2$. Then $a,b,c \in \{\pm 1, \pm 2\}$. It is clear that $a,b,c$ cannot all be equal (otherwise $a=b=c=0$) and cannot all be distinct (otherwise $a+b+c\neq 0)$. Suppose $a \geq b \geq c$. We have two cases.

(i) $a=b=1, c=-2$. This implies that $x,y,z$ are three consecutive integers, $x>y>z$. If $y=n>1$, where $n \in \mathbb{Z}$, the equation becomes
$$(n+1)^k-2n^k+(n-1)^k=2.$$ If $k=2$, we get an identity, so there are infinitely many positive integer solutions $(n+1,n,n-1)$. If $k>2$, then we have $$(n+1)^k+(n-1)^k > 2n^k+k(k-1)n^{k-2}>2n^k+2,$$ so the equation has no solution.

(ii) $a=2, b=c=-1$, so $x,y,z$ are three consecutive integers, $x>z>y$. If $z=n>1$, the equation becomes
$$-(n+1)^k-(n-1)^k+2n^k=2,$$ which gives $(n+1)^k+(n-1)^k=2(n^k-1)<2n^k$, contradiction.

In conclusion, the equation has never positive integer solutions $(x,y,z)$ if $p>3$ or $p=2, k>2$, and has infinitely many positive integer solution if $p=k=2$, namely $$(n+1,n,n-1), (n,n-1,n+1), (n-1,n+1,n), \qquad n>1.$$