Show that if p is prime and $p \nshortmid abc$ then the congruence $ax^2 + by^2 + cz^2 \equiv 0 \pmod p$ has a solution other than $(0,0,0)$

Question

Show that if p is prime and $p \nshortmid abc$ then the congruence $ax^2 + by^2 + cz^2 \equiv 0 \pmod p$ has a solution other than $(0,0,0)$

I don't see any clear path to tackle the problem. I've been trying with quadratic residues but I haven't noticed anything worthy. I've also done a lot of examples but I haven't found any method to find solutions

Answer 1

Since $p\nmid abc$, we have $$ \left|\,A =\left\{ax^2:x\in\mathbb{Z}_p\right\}\,\right| =\frac{p+1}2 $$ and $$ \left|\,B =\left\{-c-by^2:y\in\mathbb{Z}_p\right\}\,\right| =\frac{p+1}2 $$ Since $|A\cup B|=|A|+|B|-|A\cap B|$, and $|A\cup B|\le\left|\mathbb{Z}_p\right|=p$, we must have $$ |A\cap B|\ge1 $$ Therefore, as Thomas Andrews comments, there is a solution with $z=1$.