Pythagorian quadruples From my work on hyperelliptic equations I found how to get infinitely many solutions of the equation $a^4+b^4+c^2=d^4$. I call these solutions harmonic:
$$\begin{array}{rcccccl}
1^4 &+& 2^4 &+& 8^2 &=& 3^4\\
2^4 &+& 3^4 &+& 48^2 &=& 7^4\\
3^4 &+& 4^4 &+& 168^2 &=& 13^4\\
4^4 &+& 5^4 &+& 440^2 &=& 21^4
\end{array}$$
and so on. All numbers natural.
Does anyone know if there are infinite non harmonic solutions of this equation? 
 A: EDIT
There seems to be a lot of non-harmonic solutions.
Here are couple of one parameter family of non-harmonic solutions.

$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $$a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2  = (a^2 - a + 1)^2$$ You could scale these up appropriately i.e. $$\left(ka,ka(a-1),k^2(a-1)(2a^2 - a + 1),k \left(a^2-a+1 \right) \right),$$ to get other solutions.

$b = a(a+1), d = a^2 + a + 1$ and $c = (a+1)(2a^2+a+1)$ which relies on the identity $$a^4 + \left( a(a+1)\right)^4 + \left((a+1)(2a^2+a+1) \right)^2  = (a^2 + a + 1)^2$$ You could again scale these up appropriately i.e. $$\left(ka,ka(a+1),k^2(a+1)(2a^2 + a + 1),k \left(a^2+a+1 \right) \right),$$ to get other solutions.

You could also take your harmonic solution $(a,a+1,a(a+1)(a^2 + a + 2),a^2+a+1)$ and scale appropriately, i.e. $$\left(ka,k(a+1),k^2a(a+1)(a^2 + a + 2),k \left(a^2+a+1 \right) \right),$$ to get other solutions.

Yes. Below is a one parameter family of infinite solutions.
$b = a+1$, $d = a^2+a+1$, $c = (a^2+a+1)^2-1$.
$$b^4 = (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$$
$$d^4 = (a^2 + a +1)^4 = 1+4 a+10 a^2+16 a^3+19 a^4+16 a^5+10 a^6+4 a^7+a^8$$
Hence,
\begin{align}
d^4 - b^4 - a^4 & = 4 a^2+12 a^3+17 a^4+16 a^5+10 a^6+4 a^7+a^8\\
& = a^2 \left(4 +12 a+17 a^2+16 a^3+10 a^4+4 a^5+a^6 \right)\\
& = a^2 (a+1)^2 (a^2 + a + 2)^2
\end{align}
Hence, choose $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$
