Find all $(a,b,c)\in\mathbb{Z}^3$ such that $b^2-4ac=-20$, and $-|a|<b\le|a|<|c|$, or $0\le b\le|a|=|c|$.

Question

Find all $(a,b,c)\in\mathbb{Z}^3$ such that $b^2-4ac=-20$, and either of the following is true:

$-|a|<b\le|a|<|c|$, or $0\le b\le|a|=|c|$.

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We see that if $(a,b,c)$ is a solution, then so is $(-a,b,-c)$. So far I know $\pm(1,0,5)$ and $(\pm2,2,\pm3)$ work. Are these the only ones? Why or why not?

Answer 1

Hint:

1. $$-4ac\le b^2-4ac=-20\Longrightarrow ac\ge 5.$$
2. Since $|b|\le |a|\le |c|$, $$b^2-4ac=-20\Longrightarrow 4ac=20+b^2\le 20+ac\Longrightarrow ac\le 6.$$

Conclusion: What you have founded are all of the solutions of $(a,b,c)$.