Diophantine equation: $3a^2+3b^2+19ab=0$.

Question

Can this equation be solved in integers $a,b$ (Apart from $a=b=0$)? :

$3a^2+3b^2+19ab=0$

Thanks!

Answer 1

Let $(a,b)$ be a minimal nontrivial solution. Show that both $a$ and $b$ are even.

Then we get another solution $\left(\frac{a}{2},\frac{b}{2}\right)$, contradiction.

Answer 2

Let's solve the equation for $b$ as a variable (and $a$ as a constant):

• $b_{1,2}=\dfrac{-19a\pm5\sqrt{13}a}{6}=\dfrac{-19\pm5\sqrt{13}}{6}a$

• $\forall{a\in\mathbb{Z}}:a\neq0\implies{b\not\in\mathbb{Q}}\implies{b\not\in\mathbb{Z}}$

In other words, for any integer value of $a\neq0$, you get an irrational value of $b$.