A particular Diophantine equation of degree 4 in two variables

Question

Did someone met the equation of type $$a^4 + (1 - 4 b)^2 b^2 + a^2 (3 + 4 b - 16 b^2) =0$$ somewhere in practice?

I met this one in a notes on Diophantine geometry, where the equation remains unsolved.

The question there is to find at least one solution $a,b\in \mathbb{N}$ with $a>7$.

Is something known about the natural solutions of the equation?

Any suggestions on the subject are appreciated as well.

Answer 1

Idea how to aproach:

$$a^4 + (1 - 4 b)^2 b^2 + 4a^2 b(1 - 4 b)+3a^2 =0$$

Put $c=b(1-4b)$ and $d=a^2$ so we get $$d^2+c^2+4cd+3d=0$$

So $$ d^2+d(4c+3)+c^2=0$$

Now the discriminat must be perfect square:

$$ (4c+3)^2-4c^2 = e^2$$

Answer 2

Remark: for $b=2$ the equation becomes $$ (a+7)(a+2)(a-2)(a-7)=0. $$ It looks like that this Diophantine equation has been constructed, e.g., perhaps for a contest, or another challenge.

The equation is biquadratic in $a$, so we obtain that $$ 3(64b^4-32b^3-28b^2+8b+3) $$ needs to be a perfect square.