Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ I am looking for positive integer solutions to the Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$ for distinct values of $(a,b,c,d)$.
There are many solutions with $a=b$ such as $(7,7,11,13)$ and $(13,13,22,23)$, and many solutions with $c=d$ such as $(3,5,7,7)$ and $(7,8,13,13)$, but it seems not so many with $4$ distinct values.  The smallest distinct solution I know of is $(35,139,146,169)$.
Are there infinitely many solutions?  Is there perhaps a parameterization of such solutions?
Of special interest are solutions where the quantity $(c^2+d^2)-(a^2+b^2)$ is a perfect square, such as $(a,b,c,d)=(323,392,407,713)$.  Are there other solutions of this type?
This comes up in solving an unsolved problem in geometry that I can't discuss, but you might get your name on a paper if you can help.
 A: The equation you specify defines a surface $X$ in $\mathbb{P}^{3}$, and this surface is a K3 surface. It is conjectured that if $X$ is a K3 surface, there is a field extension $K/\mathbb{Q}$ over which the rational points on $X$ become Zariski dense. It's not clear that this happens over $\mathbb{Q}$.
However, one can find infinitely many solutions by finding genus $0$ curves on this surface. Specializing the polynomial to make $c = a+b$ it becomes the square of a quadratic, and this quadratic (which defines a conic in $\mathbb{P}^{2}$) has infinitely many points.
In particular, if $s$ and $t$ are positive integers with $s > t$, then
$a = s^{2}-t^{2}$, $b = 2st - t^{2}$, $c = s^{2} + 2st - 2t^{2}$, $d = s^{2} - st + t^{2}$ is a solution in positive integers to your equation. (The only solution in this family with $c^2+d^{2}-a^{2}-b^{2}$ a square is $a = c = d = 3$, $b = 0$. However, there easily could have been more.)
EDIT: The rational points on $X$ are Zariski dense. If $w$ is a rational number, specializing the polynomial at $c = aw + b$ gives a genus $1$ curve. This curve has a section, and using this, the surface $X$ is birational to the elliptic surface
$$
  E : y^{2} = x^{3} + (6w^{4} - 30w^{2} + 24)x^{2} + (9w^{8} - 90w^{6} + 189w^{4} - 144w^{2} + 36)x.
$$
It's not too hard to find two independent points of infinite order on this elliptic surface, and the preimage on $X$ of a point on $E/\mathbb{Q}(w)$ is a rational curve. In particular $X$ has infinitely many rational curves.
One of the simplest rational curves gives rise to solutions with $a < b < c < d$. In particular, let $w = \frac{s}{t}$ with $s$ and $t$ coprime positive integers and $\frac{1}{2} < \frac{s}{t} < \frac{-1 + \sqrt{13}}{4}$. Then
\begin{align*}
  a &= 32s^{5} t - 56s^{3} t^{3} + 8s^{2}t^{4} + 30st^{5} - 11t^{6}\\
  b &= -16s^{6} - 16s^{5} t + 28s^{4} t^{2} + 48s^{3} t^{3} - 31s^{2} t^{4} - 32 st^{5} + 19t^{6}\\
  c &= 16s^{6} - 16s^{5} t - 28s^{4} t^{2} + 56s^{3} t^{3} - s^{2} t^{4} - 43st^{5} + 19t^{6}\\
  d &= -16 s^{6} - 32s^{5} t + 52s^{4} t^{2} + 40s^{3} t^{3} - 47 s^{2} t^{4} - 14st^{5} + 14t^{6}
\end{align*}
is a solution with $0 < a < b < c < d$ and with $c = (s/t)a + b$. For example setting $s = 5$ and $t = 8$, each of $a$, $b$, $c$ and $d$ above are multiples of $48$ and dividing through by that gives $(1392,2197,3067,3197)$.
