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

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.

• Might be related: All solutions of the Diophantine equation $a^6+b^6=c^6+d^6+e^6+f^6+g^6$ for $a,b,c,d,e,f,g<250000$ found with a distributed Boinc project. So in your case we have $$b=f=g=0$$. Commented Feb 23, 2018 at 18:25
• This post may be of interest. Commented Feb 24, 2018 at 2:54
• @TitoPiezasIII Thanks. I've added an example which is a generalisation of the identity for $a^3+b^3=1+c^3$ in your post. Commented Feb 24, 2018 at 11:23
• You also missed $4^3 + 17^3 + 22^3 = 25^3$ for which two of the terms are square. Same with the equation, $49^3 + 84^3 + 102^3 = 121^3$ and $9^3 + 58^3 + 255^3 = 256^3$, but $9^3 + 58^3 = 22^3 + 57^3$, so does that count? Also, another solution is $118^3 + 225^3 + 324^3 = 361^3$. Commented Feb 24, 2018 at 11:40
• @user477343 Yes, there are many specific solutions of (2) with two squares - it's not hard to find them. The relevance of a parametric solution which guarantees two squares is that it opens the possibility of searching for values of the parameters that make the other two terms square too Commented Feb 24, 2018 at 11:45

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.

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.

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.

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 $$b but found no new primitive solution with $$\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.

• Commented Aug 20, 2023 at 9:02
• Thanks you for the comment i'm a bit fool to doesn't see that . BUT see projecteuclid.org/journals/… Commented Aug 20, 2023 at 11:21
• Let us start with the equation $3t-9t^4=y^2$. Multiply by $9/t^4$. We obtain the form:$$\left(\frac 3t\right)^3 -81 =\left(\frac {3y}{t^2}\right)^2\ .$$So we have a passage to the elliptic curve in Weierstraß form $Y^2 =X^3-81$, rank one, no non-trivial torsion. The generator is $P=(X_1,Y_1)=(13,46)$, let $nP$ be $(X_n,Y_n)$, so we associate $t_n=3/X_n$. The first few values for $t$ are \begin{aligned}t_1 &= 3/13 \\ t_2 &= 25392/36985 \\ t_3 &= 16007592627/48795650749 \\ t_4 &= 149981896780909248/16403197409261230945 \\ \end{aligned} Commented Aug 20, 2023 at 22:43
• So we obtain:$$-3578403985453454495^3 + 1934260992^6 + 335389956^6 = 36985^{12}\ .$$(Solution obtained inserting $t=t_2$. Note that the solution with $169^6$ on the R.H.S. above is also a "higher power", $13^{12}$. And for the next solutions...) +1 Commented Aug 20, 2023 at 22:59
• @Tito I did a brute force search, but didn't find a solution with $200 < d < 1000$. Commented Aug 22, 2023 at 1:09

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 $$15$$.

• +1 That look's promising. It implies (among other things) that $c^3$, $z^2$ and $d^3$ are in arithmetic progression, which is quite possible, eg $(c,z,d)=(1,78,23) \implies (c^3,z^2,d^3) = (1,6084, 12167)$. Commented Jul 25, 2018 at 10:32
• Good luck. It would be desirable to establish whether or not there are an infinity of solutions. Commented Jul 25, 2018 at 14:32