A system of Diophantine equation

Question

Let $a,b,c$ be three pair-wise coprime positive integers,with $ac \equiv ~1(mod ~2)$ , and $b \equiv ~0(mod~ 24)$. Consider the system of Diophantine equation

$ax+1 = c(c-a)p^2$

$bx+1 = c(c-b)q^2$

$cx+1 = (c-a)(c-b)r^2,$

where $p,q,r$ and $x$ are the variables. I have managed to show that $p,q,r$ are also pair-wise coprime with $q$ being an even number. Moreover, $x$ can be shown to be odd. My aim is to show that this system does not have any solution. Thus I seek to show my statement or disprove by providing a solution.

Answer 1

$bx+1$ is odd, but $c(c-b)q^2$ is even, contradiction.

Answer 2

Let $a=c=q=1$. Then the equations are all solved by $x=-1$ unless you also stipulate that $x$ has to be non-negative.