Diophantine Equation $a^6+b^6+c^6=d^6$

Question

Within the literature on Diophantine equations there seems to be very little on the $6,1,3$ equation:

$$a^6+b^6+c^6=d^6\quad\quad(1)$$

Mathworld, for example, simply records that there are no known solutions of $6.1.n$ equations for $n \leq 6$. The former Euler Project searching for solutions of equations in equal sums of like powers showed $6,1,5$ at the top of its most wanted list but did not mention $6,1,3$.

One way of looking at the $6,1,3$ equation is as a special case of the much more familiar $3,1,3$ equation:

$$w^3+x^3+y^3=z^3\quad\quad(2)$$

in which each of the terms is also a square. It may be considered relevant that there are known solutions of $(2)$ in which two or three of the terms are square, the smallest respectively being:

$$(1^2)^3 + 6^3 + 8^3 = (3^2)^3\quad\quad(3)$$

$$118^3 + (15^2)^3 + (18^2)^3 = (19^2)^3\quad\quad(4)$$

Question: Are there any discussions in the literature of either:

1. direct strategies for searching for solutions of $(1)$;
2. research seeking a proof that $(1)$ has no non-trivial solutions?

By a direct strategy I mean one which addresses 6,1,3 itself, rather than one which searches for $6,1,n$ for higher $n$ with the remote possibility of finding a solution in which some of the terms are zero. An example of a direct strategy would be to take the following parametric solution of $(2)$:

$$(9s^4)^3 + (3s(t^3-3s^3))^3 + (t(t^3-9s^3))^3 = (t^4)^3$$

which already has two square terms, and to try to find values of $s$ and $t$ such that the other two terms are also square.

Answer 1

COMMENT (This is not an answer!)

The identity $(2xz)^2+(2yz)^2+(z^2-x^2-y^2)^2=(x^2+y^2+z^2)^2$ gives the general solution of $X^2+Y^2+Z^2=W^2$ so one has the necessary condition with parameters $x,y,z$ $$\begin{cases}a^3=2xz\\b^3=2yz\\c^3=z^2-x^2-y^2\\d^3=x^2+y^2+z^2\end{cases}$$ This could be a good first step to get an answer.

(Follow the COMMENT).- By the way, if $$A_1=a^5+10a^4-8a^3+16a^2+64a-32\\A_2=a^5-10a^4-8a^3-16a^2+64a+32\\A_3=-a^5+8a^4+8a^3-16a^2+80a+32\\A_4=-a^5-8a^4+8a^3+16a^2+80a-32$$ then we have for all real parameter $a$ the identity $$a(6a^4+24a^2+96)^3=A_1^3+A_2^3+A_3^3+A_4^3$$ Consequently, making $a=n^3$, one has for all natural $n$ an identity:

$$X_1^3+X_2^3+X_3^3+X_4^3=X_5^3$$ where each $X_i$ has degree $15$.

Answer 2

This is related to a question of mine, but I think the answer there would be informative for this one. To find possible solutions to,

$$a^6+b^6+c^6 = d^6\tag{i}$$

one way suggested by Piquito is to look at,

$$a^2+b^2+c^2 = d^2\tag{ii}$$

and make all terms cubes. Turns out that, except for a single term (so close!), we can do so infinitely using elliptic curves. A second way suggested by the OP is to look at,

$$a^3+b^3+c^3 = d^3\tag{iii}$$

and make all terms squares. Turns out that, again except for a single term, we can do so infinitely still using elliptic curves. Examples for both are,

$$\quad 42^6+9^{12}+10^{12} = 1134865^2$$ $$25402^3+3^{18}+138^6 = 13^{12}$$

(It's curious how high powers are involved!) For the first approach, see this MSE post by Old Peter. For the second approach, consider the well-known identity,

$$(1-9t^3)^3+(3t^2)^6+(3t-9t^4)^3=1$$

To address the OP's question, we have to set both the first and third terms as squares. That might be difficult, but perhaps it is feasible for just one. Choosing the third term,

$$3t-9t^4 = y^2$$

which, after transforming to Weierstrass form, is an elliptic curve and has infinitely rational points. The first point is $t_1=3/13$ which yields the second equality below,

$$\color{blue}{118^3 + 15^6 + 18^6 = 19^6}\tag1$$ $$25402^3+27^6+138^6 = 169^6\tag2$$

where $169 = 13^2$ and are the 1st and 2nd smallest primitives with $\text{GCD}(b,c,d)=1.$ (I did a small exhaustive search with $d<200.$) Another solution is $t_3 = \frac{(73047\sqrt{3})^2}{48795650749}$, plus an infinite more, though there may be points of smaller height.

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Edit 1: Thanks to Dan Fulea who found the smaller height $t_2 = \frac{(92\sqrt{3})^2}{36985}$ which yields,

$$\small{−3578403985453454495^3+1934260992^6+335389956^6=1367890225^6}\tag3$$

and $1367890225=36985^2.$ Whether these 3 solutions are the smallest primitives with $d<10^{10}$ is rather doubtful, since the first solution is sporadic, does not belong to the infinite family, and there may be more sporadics.

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Edit 2: All of Dan's seven $t_n$ in the comments below have the form $t_n = \frac{(p\sqrt{3})^2}{q}$. If this is always the case, then,

$$\; -a+d^2 \equiv 0 \text{ mod }3^5$$

But this is also obeyed by the first sporadic solution $ -118+19^2 = 3^5.$ So whether this is a valid congruence for all primitive $a^3+b^6+c^6 = d^6$ is unknown, and whether it has other sporadic solutions (like the blue equation) is also unknown.

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Edit 3: Thanks to Seiji Tomita for extending the search radius! He searched for integer $a=(-b^6-c^6+d^6)^{1/3}$ with $b<c<d<1000$ but found no new primitive solution with $\text{GCD}(b,c,d)=1.$ Still, there is large gap between $d=10^3$ and $d=10^{10}$, so there might be sporadic solutions in between.