Small Galois group solution to Fermat quintic I have been looking into the Fermat quintic equation $a^5+b^5+c^5+d^5=0$. To exclude the trivial cases (e.g. $c=-a,d=-b$), I will take $a+b+c+d$ to be nonzero for the rest of the question. It can be shown that if $a+b+c+d=0$, then the only solutions in the rational numbers (or even the real numbers) are trivial.
I had the idea of representing the values $a, b, c, d$ as roots of a quartic polynomial, and then trying to force the Galois group to have low order. As the Fermat quintic is symmetric, it yields an equation in the coefficients of the quartic, which are satisfied by the quartic equations: $5x^4-5x^3+5qx^2-5rx+5(q-1)(q-r)+1=0$, so clearly the most general case of Galois group $S_4$ is attainable. One could also fix one or two of the roots, obtaining a smaller Galois group.
However, what I have yet to discover, is a quartic in the family above which has a square discriminant, forcing the Galois group to be a subgroup of the alternating group $A_4$. Is there an example of such a quartic? Is there one with Galois group a proper subgroup of $A_4$? Better yet, are there infinite families of such quartics?

As I understand the question, the OP wishes to solve,
$$x_1^k + x_2^k = x_3^k + x_4^k\tag1$$
for $k = 5$ and where the $x_i$ are roots of quartics. If we allow that the $x_i$ are roots of different quartics, then there are in fact solutions for $k = 5,6,8$.

I. k = 5
The first identity below just uses quadratics,
$$(a\sqrt2+b)^5+(-b+c\sqrt{-2})^5 = (a\sqrt2-b)^5+(b+c\sqrt{-2})^5$$
with Pythagorean triples $(a,b,c)$ and was known by Desboves. The second is by yours truly,
$$(\sqrt{p}+\sqrt{q})^5+(\sqrt{p}-\sqrt{q})^5 = (\sqrt{r}+\sqrt{s})^5 + (\sqrt{r}-\sqrt{s})^5$$
where,
\begin{align}
p &= 5vw^2,\quad q = -1+uw^2\\
r &= 5v,\quad\quad s = -(u+10v)+w^3\end{align}
and $w = u^2+10uv+5v^2$.

II. k = 6
We use the same form,
$$(\sqrt{p}+\sqrt{q})^6+(\sqrt{p}-\sqrt{q})^6 = (\sqrt{r}+\sqrt{s})^6 + (\sqrt{r}-\sqrt{s})^6$$
where,
\begin{align}
p &= -(a^2+14ab+b^2)^2+(ac+bc+13ad+bd)(c^2+14cd+d^2)\\
q &= \;\;(a^2+14ab+b^2)^2-(ac+13bc+ad+bd)(c^2+14cd+d^2)\\
r &= \;\;(c^2+14cd+d^2)^2-(ac+13bc+ad+bd)(a^2+14ab+b^2)\\
s &= -(c^2+14cd+d^2)^2+(ac+bc+13ad+bd)(a^2+14ab+b^2)
\end{align}

III. k = 8
Still using the same form,
$$(\sqrt{p}+\sqrt{q})^8+(\sqrt{p}-\sqrt{q})^8 = (\sqrt{r}+\sqrt{s})^8 + (\sqrt{r}-\sqrt{s})^8$$
where,
\begin{align}
p &= n^3-2n+1\\
q &= n^3+2n-1\\
r &= n^3-n-1\\
s &= n^3-n+1\end{align}
P.S. Unfortunately, $k= 7$ and $k=9$ does not seem to be amenable to the same approach.
