Let $p$ be a prime number. How mane triples of integers $(x,y,z)$ satisfies $p(xy+xz+yz-p(x+y+z))=xyz$

Question

Let $p$ be a prime number. How many triples of integers $(x,y,z)$ satisfies the ecuation $$p(xy+xz+yz-p(x+y+z))=xyz?$$

Can someone help me with this problem?

I've been trying to solve it for few hours without any result.

Answer 1

Write the equation as $$-(x+y+z)p^{2} + p(xy+xz+yz) - xyz = 0$$ By Vieta's formulas, this is $$(p-x)(p-y)(p-z)-p^{3}=0 \implies (p-x)(p-y)(p-z)=p^{3}$$ The only decompositions of $p^{3}$ into positive integers are (permutations of) $(p^{2},p,1)$, $(p^{3},1,1)$, or $(p,p,p)$. These are all different. So there are $6+3+1=10$ choices of $(x,y,z)$ for each prime $p$.

As @Wojowu points out, there are additional choices from negating two of the three factors. The numbers $\pm p, \pm p^{2}, \pm p^{3}, \pm 1$ are all different if $p$ is a prime number. There are 3 distinct ways to do this in the first case, 2 in the second case and 1 in the last case, if I see right. So I think the correct answer is actually $6*3+3*2+1*1=25$. (I may have miscounted again, if so just let me know and I'll think harder).

EDIT: After some numerical checking, I find my calculations are still not careful enough. The solutions outlined above for the triple $(p-x,p-y,p-z)$ are the following, with the corresponding values of $(x,y,z)$ on the right: \begin{align*} (p^{2} ,p ,1 ) &\implies (p-p^2,0 ,p-1) \\ (-p^{2},-p,1 ) &\implies (p+p^2,2p ,p-1) \\ (p^{2} ,-p,-1) &\implies (p-p^2,2p ,p+1) \\ (-p^{2},p ,-1) &\implies (p+p^2,0 ,p+1) \\ (p^{3} ,1 ,1 ) &\implies (p-p^3,p-1,p-1) \\ (-p^{3},1 ,-1) &\implies (p+p^3,p-1,p+1) \\ (p^{3} ,-1,-1) &\implies (p-p^3,p+1,p+1) \\ (p ,p ,p ) &\implies (0 ,0 ,0 ) \\ (p ,-p,-p) &\implies (0 ,2p ,2p ) \end{align*}

So we need to check when the following numbers are pairwise equal: $p\pm p^{2}$, $p \pm p^{3}$, $p\pm 1$, $2p$, $0$. They are always distinct. Therefore, by permuting the elements in each triple, we get

$$6+6+6+6+6+3+1+3+3 = 40.$$ (I've checked this by computer for $p=2,3$).