Solutions to $a^2+b^2+(ab)^2=c^2$ In a comment to the question $n^2+(n+1)^2+n^2\cdot(n+1)^2$ is a perfect square it is proved that:
$(1)\quad b=a+1$ give integer solutions to 
$a^2+b^2+(ab)^2=c^2$ for all $a\in\mathbb N$.
In the answers to the question $\forall m\in\mathbb N\exists n>m+1\exists N\in\mathbb N:m^2+n^2+(mn)^2=N^2$ it is proved that:
$(2)\quad b=2a^2$ give integer soulutions for all $a\in\mathbb N$.
Conjecture:

For each $a\in\mathbb N$ there are integer solutions to
  $a^2+b^2+(ab)^2=c^2$ that are neither of the types $(1)$ or $(2)$
  above.

Prove the conjecture or give a counter-example. It is tested for all $a<1000$.
 A: The Pell type equation $$ x^2 - (1 + A^2) y^2 = A^2  $$ has an infinite set of solutions $(x,y).$ The first three predictable types are
$$
\left(
\begin{array}{c}
A \\
0
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 - A + 1 \\
A - 1
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 + A + 1 \\
A + 1
\end{array}
\right).
$$
After that, an infinite sequence of $(x,y)$ as column vectors may be found by multiplying by the generator of the (oriented) automorphism group of the binary quadratic form $ x^2 - (1 + A^2) y^2. $ That matrix is
$$ M =
\left(
\begin{array}{cc}
2A^2  + 1 & 2 A^3 + 2 A \\
2A  & 2 A^2 + 1
\end{array}
\right).
$$
The answer you write as $b = 2 a^2$ comes up as
$$ M =
\left(
\begin{array}{cc}
2A^2  + 1 & 2 A^3 + 2 A \\
2A  & 2 A^2 + 1
\end{array}
\right)
\left(
\begin{array}{c}
A \\
0
\end{array}
\right) =
\left(
\begin{array}{c}
2A^3 + A \\
2 A^2
\end{array}
\right).
$$
The bad news is that, for a fixed $A,$ there may be others. These others show up as later entries in the Pell sequences for smaller values of $A.$ I do not have a genuinely sensible two-dimensional description for all answers.
