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?
 A: I have answers to your first two questions, and some insight into the third.
First, there is a quartic in the family above with Galois group contained in $A_{4}$. One example is found by taking $q = 1/5$ and $r = 0$. This gives $5x^{4} - 5x^{3} + x^{2} + 1/5$, which has Galois group $A_{4}$. A quartic has Galois group contained in $A_{4}$ if and only if the discriminant $D(q,r)$ is a square, and the discriminant of your quartic family is degree $4$ in $r$. This means that if you find a single $A_{4}$ polynomial in the family, this gives a rational point on $y^{2} = D(q,r)$. This is a genus $1$ curve, and there's a good chance that this gives you an elliptic curve with rank $\geq 1$ and hence infinitely many rational $r$ with the same $q$. In particular, for $q = 1/5$, the elliptic curve is $y^{2} + xy = x^{3} + x^{2} - 12642x - 612239$ and has rank $2$.
Second, there are quartics in your family with Galois group a proper subgroup of $A_{4}$, namely $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. I know of three examples: $(q,r) = (49/120,1/12)$, $(791/1560,17/120)$, and $(1949/3680,189/736)$. In the first case, the splitting field is $\mathbb{Q}(\sqrt{-3},\sqrt{10})$. In the second case, the splitting field is $\mathbb{Q}(\sqrt{10}, \sqrt{-1599})$. In the third case, the splitting field is $\mathbb{Q}(\sqrt{-230},\sqrt{410})$.
Third, I do not know if there are infinite families of such quartics. I focused on the case that the Galois group is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. This occurs when the resolvent cubic has a rational root, and when the discriminant is a square. The resolvent cubic in your family is a cubic curve in $x$, $q$ and $r$ and a projectivization of this polynomial defines a cubic surface in $\mathbb{P}^{3}$. This surface is rational (and so there is a two parameter subfamily of your family where the Galois group is contained in the dihedral group of order $8$). Imposing the restriction that the discriminant is a square leads to a surface of the form $X : y^{2} = f(a,b,c)$, where $f$ is a homogeneous degree $8$ polynomial. There are a couple of rational curves on $X$ that arise from the points at infinity on the projectivization of the resolvent cubic, which do not correspond to pairs $(q,r)$. I have not found any other curves of genus $0$ or $1$ on $X$. A computation with Magma indicates that $X$ is a K3 surface, and it is conjectured that there is a number field $K$ for which $X(K)$ is Zariski dense. It is plausible that $X$ has many rational points.
A: Here is a construction that appears to give a 2-parameter family of solutions that lie to cubic extensions of $\mathbb{Q}$. The construction uses some basic algebraic geometry.
Consider the line $L=V(a+b,c+d)$ in the projective surface $S=V(a^5+b^5+c^5+d^5)$ in $\mathbb{P}^3$.
For each point $p$ of $L$ consider the (projective) tangent plane $P_p=\overline{T_p(S)}$ to $S$ at $p$; this plane contains $L$.
The intersection of $P_p$ and $S$ is the union of $L$ and a quartic curve $C_p$ in the projective plane $P_p\cong\mathbb{P}^2$, which has a point $p$. Projecting from this point gives a map $C_p\to\mathbb{P}^1$ of degree $3$.
Thus, choosing a point $t$ of $\mathbb{P}^1$ gives an equation of degree $3$ over $\mathbb{Q}$ whose solutions give points on the curve $C_p$ and hence on the surface $S$. The choice of $p$ and $t$ give two parameters for this system of equations.
The analogous construction for $a^3+b^3=c^3+d^3$ can be found in an article in "Resonance".
A: 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.
