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

Related/same proof technique: Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p\mid x^{2} + y^{2} + 1$. — Bart Michels, Oct 16 '16 at 21:57
I am sure the technique in the thread Barto linked has been used way earlier. But, this is the oldest hit I could find. Weak search-fu :-( — Jyrki Lahtonen, Oct 16 '16 at 22:30
@JyrkiLahtonen I guess you mean on this site; but it is a step in the method Hardy and Littlewood use to show the sum of four squares is positive universal — Will Jagy, Oct 16 '16 at 22:50

score 3 · Accepted Answer · edited Oct 16 '16 at 23:18

3

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 $z=1$.

edited Oct 16 '16 at 23:18

darij grinberg

17,777

answered Oct 16 '16 at 22:30

robjohn

345,667

Did you mean $B = {by^2+c}$? Also, couldn't the intersection between both of them end up implying that x and y equal zero? – McNuggets666 Oct 16 '16 at 22:39
$B = {by^2+c}$ – McNuggets666 Oct 16 '16 at 22:40
1

@CamiloEscobar: no, I meant $B =\left{-c-by^2:y\in\mathbb{Z}_p\right}$. That way, any element of $A\cap B$ says that $ax^2\equiv-c-by^2\pmod{p}$. – robjohn Oct 16 '16 at 22:41
@CamiloEscobar: as for $x=y=0$, since $z=1$, we won't have $(x,y,z)=(0,0,0)$. – robjohn Oct 17 '16 at 00:19

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 $(0,0,0)$

1 Answers1