Please help me generate new solutions from $40425^6+45990^6+40802^6=135794767970233^2$

As far as I’m aware, I’ve found the only known primitive solution to $x^6+y^6+z^6=w^2$ with $7|x$ and $7|y$, and I would like to generate further solutions from this one. That solution is

$$40425^6+45990^6+40802^6=135794767970233^2$$

I’m especially interested to see if the${\pmod 7}$ properties are preserved in any generated solutions.

My question

Can this solution be used to produce other non-trivial solutions?

If so, what are those solutions, and, in the simplest terms, how are they found.

Remarks

$$7^63^6|w+z^3$$ $$5^6|w-z^3$$ $$40425=105\cdot385$$ $$45990=105\cdot438$$

These links are the reason I suspect there might be family of solutions

http://www.maroon.dti.ne.jp/fermat/dioph149e.html

http://www.maroon.dti.ne.jp/fermat/dioph150e.html

but I’ve not been able the follow the method.

These are my earlier relevant questions

The Diophantine equation $x_1^6+x_2^6+y^6=z^2$ where both $(x_i)\equiv 0{\pmod 7}$.

The Diophantine equation $x_1^6+x_2^6+x_3^6=z^2$ where exactly one $(x_i)\equiv 0{\pmod 7}$.

As I’m very far from my comfort zone with this, I apologise in advance if this is a stupid question.

• Any integer like $n^6$ is a perfect square too. so multiplying both sides of your relation by $n^6$ gives new solution. Commented May 17, 2018 at 14:42
• No, I don’t think it does, but by all means show me a numerical example, I could be wrong. Although $(kx,ky,kz,k^3w)$ does give more solutions, none are primitive. Thanks for your interest. Commented May 17, 2018 at 15:55
• @OldPeter You may be interested in this post. Your case and that case both involve a four-term equality where three terms are $6$th powers. And both versions have infinitely many primitive solutions by solving an elliptic curve. Commented Aug 22, 2023 at 5:16

Part I.

I'm very late for this party but, yes, there are infinitely many solutions to,

$$(7a)^6 + (7b)^6 + c^6 = d^2$$

such as,

$$\small{8087694419216299774939070914834367221515446235961718559433991242747679355^6 \\ + 39981928503165529971434133622219927093673635787864763118449995966157311010^6 \\ + 36592238812351494777854101136491581880454567677200804220398933530326048862^6 \\ = d^2}$$

where $$d$$ has about 220 decimal digits.

Part II.

As requested by the OP, here is the procedure.

We use the clever method by Bremner and Ulas in a 2011 paper. Essentially, what they did for $$a^6+b^6+c^6 = d^2$$ was similar to what Elkies did for $$a^4+b^4+c^4 = d^4$$: namely, break a Diophantine quartic into two quadratics, hence it becomes an intersection of two quadric surfaces. For the sextic case, we assume,

$$x^6 + y^6 + z^6 = d^2 = (ux^2 + uy^2+z^3)^2$$

to get the quartic in $$(x,y)$$,

$$P := x^4 - u^2 x^2 - u^2 y^2 - x^2 y^2 + y^4 - 2 u z^3$$

If $$P=0$$, and solving for $$u$$,

$$u = \frac{-z^3+\sqrt{x^6+y^6+z^6}}{x^2+y^2}$$

so it seems we are back to where we started. But they found $$P$$ can be expressed in terms of quadric surfaces as,

$$P:=\frac{e_1 e_2}{(a+b)^2}+\frac{f_1 f_2}{(a+b)^2}$$

for appropriate constants $$(a,b)$$ and where,

\begin{align} e_1 &= a(u^2 + x^2 - y^2 + z^2) + b(t + u^2 - y^2)\\ e_2 &= a(-y^2 - z^2) + b (-t + x^2 - y^2)\\ f_1 &= a^2 (t - 2u z + x^2 + z^2) + a b (2t - 4u z - u^2 + x^2) - b^2 (2u z + u^2 - x^2)\\ f_2 &= -t + x^2 + z^2\end{align}

with $$t = u x - x z + u z.$$

Part III.

Notice that $$(e_k, f_k)$$ only involve $$2$$nd powers. So if $$e_1 = f_1 = 0$$ (or $$e_1 = f_2 = 0$$), then $$P=0$$ and we have a solution. However, we already have the OP's solution (with $$y,z$$ switched),

$$(x,y,z) = (40425, 40802, 45990)$$

We plug the three into $$u =\dfrac{-z^3+\sqrt{x^6+y^6+z^6}}{x^2+y^2} \to 11677$$, then the four into $$e_1$$, and we get $$a = 16981$$, $$b = 15855$$. Pluging $$(a,b)$$ back into $$(e_1, f_1)$$ now with $$(u,x,y,z)$$ as indeterminates, we get,

$$R_1 := 32836 u^2 + 16981 (x^2 + z^2) + 15855 (u x - x z + u z) = 32836 y^2$$ $$R_2 := -520614780 u^2 + 808969141 x^2 + 288354361 z^2 + 826821871 (u x - x z) - 1329583921 u z = 0$$

which must be solved simultaneously. $$R_1$$ has 4 variables, while $$R_2$$ has only 3 so the latter will be solved first. But to generate an elliptic curve, $$R_2$$ has to be solved parametrically.

Part IV.

Using the OP's $$(x,z)$$, the three variables $$(u,x,z)$$ of $$R_2$$ can be solved parametrically as,

$$u = -162639678860914069753177 p^2 + 221687867195393559497924 p q - 77983014642252680812447 q^2$$ $$x = \color{blue}{7\times15}\, (805382988663926749091 p^2 - 2074612904930049947436 p q + 911787635157613275985 q^2)$$ $$z = \color{blue}{7\times30}\, (518653226232512486859 p^2 - 911787635157613275985 p q + 236039342456178049276 q^2)$$

with the last two as multiples of $$7$$ as requested. What remains is to find the last variable $$y$$.

Part V.

We substitute the polynomials $$(u,x,z)$$ above into $$R_1$$ to get,

$$Poly(p,q) = 32836y^2$$

where $$Poly(p,q)$$ is only a quartic. Since the OP's solution in an initial rational point, we then know this is birationally equivalent to an elliptic curve with another point as,

$$p = 789725624211424257106066446522166922409\\ q = 1495397915759421270794173585425020145620$$

and infinitely more $$(p,q)$$.

P.S. I used the tangent-chord method so there might be smaller points. And of course, one has to remove common factors to get the primitive solution in Part 1.

• Thank you for your answer; what an encouraging surprize! I would love to see a few more details, perhaps the equation or any smaller solutions, but I’m still pleased if you can’t. Commented Feb 6, 2023 at 19:21
• @OldPeter Details have been added. Commented Feb 7, 2023 at 15:22
• Thank you so very much, Tito, for spending so much of your time on this. Commented Feb 7, 2023 at 19:25
• @OldPeter If you have the computing power and time, you may want to check out this MO post. Commented Feb 15, 2023 at 6:43
• @OldPeter Oops, sorry. Name has been deleted. Commented Sep 17, 2023 at 11:03