When a quadratic involving three primes is a perfect square

Question

How do we find all primes $p,q,r$ such that $p^2+q^2+rpq$ is a perfect square ?

with $r=7$ and $p=q$ we have the expression a perfect square

Answer 1

Generally before searching primes need to understand what formula solutions of this equation. Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.

Write the formula can someone come in handy. the equation:

$Y^2+aXY+X^2=Z^2$

Has a solution:

$X=as^2-2ps$

$Y=p^2-s^2$

$Z=p^2-aps+s^2$

more:

$X=(4a+3a^2)s^2-2(2+a)ps-p^2$

$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$

$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$

more:

$X=(a+4)p^2-2ps$

$Y=3p^2-4ps+s^2$

$Z=(2a+5)p^2-(a+4)ps+s^2$

more:

$X=8s^2-4ps$

$Y=p^2-(4-2a)ps+a(a-4)s^2$

$Z=-p^2+4ps+(a^2-8)s^2$

For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.

$X=3s^2+2ps$

$Y=p^2+2ps$

$Z=p^2+3ps+3s^2$

more:

$X=3s^2+2ps-p^2$

$Y=p^2+2ps-3s^2$

$Z=p^2+3s^2$

In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.

$X=s^2-bp^2$

$Y=ap^2+2ps$

$Z=bp^2+aps+s^2$

$p,s$ - integers asked us.