Finding sixth powers like $65^6 + 52^6 + 15^6 = 36^6 + 67^6 + 37^6$ using Pythagorean quadruples $65^2 = 52^2 + 15^2 + 36^2$?

Question

I. Quadruples

In this post, we saw how infinitely many Pythagorean triples such as,

$$3^2+4^2-5^2=0\quad$$

can lead to 4th power equalities,

$$2^4+2^4+4^4+3^4+4^4=5^4\quad$$

It turns out that infinitely many Pythagorean quadruples such as

$$\color{blue}{-65^2 + 52^2 + 15^2 + 36^2} = 0\qquad$$

can now lead to 6th power equalities,

$$\color{blue}{65^6 + 52^6 + 15^6} = \color{blue}{36^6} + (52 + 15)^6 + (52 - 15)^6$$

which is the second smallest $(6,3,3)$ known. Or another example,

$$\color{red}{-10947^2 + 10591^2 + 902^2 + 2618^2} = 0\qquad$$

leading to,

$$\color{red}{10947^6 + 10591^6 + 902^6} = \color{red}{2618^6} + (10591 + 902)^6 + (10591 - 902)^6$$

and so on, though the numbers get large fast since an elliptic curve is involved. As,

$$x_1^6+x_2^6+x_3^6 = y_1^6+y_2^6+y_3^6\qquad$$

they also obey,

$$x_1y_1(x_1^2-y_1^2)+x_2y_2(x_2^2-y_2^2)+x_3y_3(x_3^2-y_3^2)=0$$

hence are actually three right triangles whose areas (signed) sum to zero.

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II. Elliptic curve

Consider the system,

\begin{align} a^2+b^2+c^2 &= d^2+(b+c)^2+(b-c)^2\\[5pt] a^6+b^6+c^6 &= d^6+(b+c)^6+(b-c)^6 \end{align}

This was first discussed in a 1974 paper by Brudno and Kaplansky. They used some complicated substitutions and didn't seem to notice that the first equation reduced to the simple quadruple $-a^2+b^2+c^2+d^2=0$. Using that fact, we can more quickly find a solution to the quadratic-sextic system as,

\begin{align} c &= \frac{a\sqrt{a^2-b^2}}{\sqrt{a^2+9b^2}}\\[6pt] d &= \frac{3b\sqrt{a^2-b^2}}{\sqrt{a^2+9b^2}} \end{align}

or to solve,

$$(a^2-b^2)(a^2+9b^2)=y^2$$

If it has a rational point, then a quartic to be made a square is birationally equivalent to an elliptic curve, hence there are infinitely many integer $(a,b)$ though avoiding the trivial $(a^2-b^2)(a^2-9b^2)(4a^2-9b^2)b^2=0$.

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III. Question

Is is true that solutions to, $$(a^2-b^2)(a^2+9b^2)=y^2$$ come in unsigned pairs which yield the same $x_1^6+x_2^6+x_3^6 = y_1^6+y_2^6+y_3^6$ and one of the pair has $a+b=z_1^2$ while the other has $a-b=z_2^2$ ?

Example:

1st pair: $(a_1,b_1)=(13,3)$ where $13+3=4^2$ and $(a_1,b_1)=(5,4)$ where $5-4=1^2$.

2nd pair: $(a_2,b_2)=(267,22)$ where $267+22=17^2$ and $(a_2,b_2)=(123, 119)$ where $123-119=2^2$.

P.S. What are the 3rd, 4th, 5th, and do they obey the same?

Answer 1

$$ (a^2-b^2)(a^2+9b^2)= v^2 $$
Let $x=\dfrac{a}{b}, y= \dfrac{v}{b^2}$. We have

$$y^2 = x^4+8x^2-9 \tag{1}$$

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I. First elliptic curve

This equation can be transformed to Weierstrass form with $(X,Y)=(x^2,xy)$ $$E_1: Y^2=X^3+8X^2-9X$$ $E_1$ has has rank $1$ and generator is $P(X,Y)=(-4,10).$
We get the solutions using the group structure $kP$ with $k=1\cdots10$ as follows.

Since the solutions are huge, we only show $(a,b,a-b,a+b).$

(2929, 1560, 37^2, 67^2)  
(112982885, 19106164, 9689^2, 11493^2)
(126901625119681, 125074345690320, 1351769^2, 15873751^2)
(23501593203523539455525, 6983371409262733246316, 128523234453^2, 174599440471^2)

Remark: The system of equations $(a^2-b^2=u^2,a^2+9b^2=v^2)$ is reduced to $y^2=10x^4-16x^2+10$ and it is birationally equivalent to $E_3: Y^2=X^3-X^2-30X+72.$ The previous $E_1: Y^2=X^3+8X^2-9X$ is also birationally equivalent to $E_3.$ This is why $E_1$ gives the solutions such that $a^2-b^2$ is square.

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II. Second elliptic curve

Since the equation $(1)$ has a solution $(x,y)=(1,0)$, it is birationally equivalent to another elliptic curve

$$E_2:Y^2=X^3 -X^2 +15X + 225$$

yielding $(a,b,a-b,a+b)$

(3, 2, 1, 5)  
(13, 3, 10, 4^2)
(267, 22, 245, 17^2)
(617307, 564262, 53045, 1087^2)
(126690733, 111355677, 3916^2, 238046410)

where $(a_0,b_0) = (3,2)$ is trivial but continues to non-trivial solutions $(a_1,b_1) = (13, 3)$ and $(a_2,b_2) = (267, 22)$ mentioned in the post by Piezas.