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This post had an answer that involved the equation,

$$x^4+y^4+z^4 =t^2$$

Holding $y,z$ constant, this is then a quartic polynomial in $x$ to be made a square. Using a birational transformation, it can be transformed into an elliptic curve. So given an initial rational point $x_0$, one can find an infinite more.

For example, given the Pythagorean triple $a^2+b^2=c^2$, Fauquembergue's elegant solution is,

$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2\tag1$$

From this, I got a second but not so elegant solution,

$$x = \frac{-2a^4b^4+d^2}{2abd}, \quad y =ac,\quad z =bc\tag2$$

where $d = a^4+a^2b^2+b^4$. Scaling variables gives integer $x,y,z$.

However, I used a short-cut, the tangent method, so I'm not sure if $(2)$ is the next simplest. (The first yields $t$ that is $4$th deg in $a,b$, while the second has $t$ that is $16$-deg.)

Q: Is there a simpler solution $x$ such that $t$ as a polynomial in $a,b$ has deg $4<k<16$?

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  • $\begingroup$ $$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$ Formulate the problem differently. To solve such a system. $2ps=a^2$ ; $2ks=b^2$ ; $p^2+k^2-s^2=c^2$ $\endgroup$ – individ Jan 1 '18 at 14:26

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