Find all $x,y,p \in \Bbb Z$ with $p$ being a prime and $x^2-3xy+p^2y^2=12p.$

Question

Find all $x,y,p \in \Bbb Z$ with $p$ being a prime and $$x^2-3xy+p^2y^2=12p.$$

Looking at this mod $3$ one has that $$x^2+p^2y^2 \equiv0 \pmod{3}.$$

Since any square mod $3$ is either $0,1 $ we have that $x^2$ and $p^2y^2$ are both congruent to $0$ mod $3$. This would imply also that since they're both multiples of $3$ and squares we have that $x^2,p^2y^2 \equiv0 \pmod{9}.$ Thus $12p \equiv 0 \pmod{9} \Rightarrow p = 3.$

Now we have $$x^2-3xy+9y^2-36=0.$$

My idea was to use the quadratic formula here since we want integer solutions that might be something to look at. I got that $$x=\frac{3xy\sqrt{9x^2y^2-4(9y^2-36)}}{2}.$$

Now for integer solutions, the term inside the rad must be a perfect square right?

Solving for $9x^2y^2-36y^2+144=0 \Rightarrow x^2y^2-4y^2+16=0 \Rightarrow x^4-4=-\frac{16}{y^2}.$

So $y^2$ has to divide $16$, would imply that $y \pm2.$ For these $y$ values I get that $x=0.$

So possible solutions are $(0,2,3),(0,-2,3)$. However there seems to be more solutions still... Altogether there's six in total I think. What might I be missing here?

Answer 1

For $p=3$ the equation is equivalent to $$ (3y-x)x = 9(y + 2)(y - 2) $$ But for both cases, $y=2$ or $y=-2$ you name, we obtain two solutions for $x$, and not just $x=0$.

Moreover you missed the two solutions with $y=0$.

So indeed, altogether we have six solutions. We do not have more, since $16-3y^2$ must be a square, which is only true for $y=0,y=2,y=-2$.

Answer 2

I'll start from your equation $x^2 - 3xy + y^2 = 36$ and go in a different way. We can complete the square to get

$$(x-\tfrac{3}{2}y)^2 + \tfrac{27}{4}y^2 = 36$$

The LHS is a sum of two squares. So certainly $y^2 < 16/3 < 9$, i.e. $|y| < 3$.

This gives us four possible values for $(x-\tfrac32y)^2$, namely:

$$36, 117/4, 9, -99/4$$

We can obviously rule out $117/4$ (not a square) and $-99/4$ (negative). So $y=0, \pm2$.

• If $y=0$, we get $x^2 = 36 \implies x = \pm 6$.
• If $y=2$, we get $(x-3)^2 = 9 \implies x = 0,6$.
• If $y=-2$, we get $(x+3)^2 = 9 \implies x = 0,-6$.

Altogether, that makes six solutions.