Is the equation $a^4+b^4+c^4=2d^2$ with $a=b\,$ completely solved by $x^4-2y^4=\pm z^2$?

Question

In this answer, to solve $a^4+a^4+b^4 = 2d^2$ the author uses a nice recursion. Assume primitive solutions with $\text{gcd}(a,b)=1$.

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I. Recursion 1

Starting with,

$$1^4+1^4+2^4 = 2\times\color{blue}{3}^2\tag1$$

Let $p=\color{blue}{3},\,q=1\times2.\,$ Since $\sqrt{p^4-2q^4\,}=7,\,$ and $2pq=12$, then,

$$7^4+7^4 +12^4 = 2\times\color{blue}{113}^2\tag2$$

Let $p=\color{blue}{113},\,q=7\times12.\,$ Since $\sqrt{p^4-2q^4\,}=7967,\,$ and $2pq=18984$, then,

$$7967^4+7967^4+18984^4 =2\times\color{blue}{262621633}^2\tag3$$

Let $p=\color{blue}{262621633},$ and so on, with the numbers getting large fast. The structure of the equation used the primary form,

$$r^4 + r^4 + (2p q)^4 = 2(p^4 + 2 q^4)^2$$

where $p^4-2q^4=r^2$. Note that a larger solution,

$$(p^4 + n q^4)^4 - n(2p q r)^4 = (4n p^4q^4 - r^4)^2$$

can come from a smaller solution to $p^4-nq^4=\pm r^2$, explaining part of the recursion. However, it missed a smaller one,

$$ 239^4 + 239^4 + 26^4 = 2\times57123^2\tag4$$

pointed out by Empy2 in the comments of that post.

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II. Recursion 2

I realized the two factors of the primary form, namely $p^4-2q^4=\pm r^2$, were in fact both useful and solvable, and explained the missing solution. Thus, $p^4-2q^4=-r^2$ has $(p_0,q_0)=(1,1)$ and $(p_1,q_1)=(1,13)$.

Let $p=1,\,q=13.\,$ Since $\sqrt{p^4-2q^4\,}=239i,\,$ and $2pq=26$, then,

$$239^4 + 239^4 + 26^4 = 2\times\color{blue}{57123}^2\tag4$$

Let $p=\color{blue}{57123},\,q=239\times26.\,$ Since $\sqrt{p^4-2q^4\,}=3262580153,\,$ and $2pq=709924644$, then,

$$3262580153^4 + 3262580153^4 + 709924644^4 = 2\times10650393355715621873^2\tag5$$

and so on.

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III. Recursion 3

However, Tomita found a second point $(p_2,q_2)=(1343, 1525)$.

Let $p=1343,\,q=1525.\,$ Since $\sqrt{p^4-2q^4\,}=2750257i,\,$ and $2pq=4096150$, yielding a smaller one,

$$2750257^4 + 2750257^4 + 4096150^4 = 2\times\color{blue}{14070212996451}^2\tag6$$

He did a search and there were no other $(p_n,q_n)<10^6$. In fact, using elliptic curves, he found these 6 are the smallest primitives within a bound.

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IV. Question

Are all solutions to $a^4+a^4+b^4 = 2d^2$ given by the primary form, $$r^4 + r^4 + (2p q)^4 = 2(p^4 + 2 q^4)^2$$

where $p^4-2q^4=\pm r^2$, or are there some not of that form?

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Update:

A. Solutions to $p^4-2q^4=-r^2$

$p = 1, 1343, 2372159, 9788425919,\dots$ (A167437, Integers $k$ such that $a + b = k^2$ and $a^2 + b^2 = m^4$)

$q = 13, 1525, 2165017, 42422452969,\dots$ (A166929, Integers $m$ such that $a + b = k^2$ and $a^2 + b^2 = m^4$)

Note: The problem of finding a Pythagorean triple $(a,b,c)$ such that $a+b$ and hypotenuse $c$ are both squares was discussed by Fermat and Mersenne (see A166930). For positive $(a,b)$, Fermat found the smallest has $q=m=2165017$,

$$4565486027761+ 1061652293520 =2372159^2$$ $$4565486027761^2+ 1061652293520^2 =2165017^4$$

B. Solutions to $p^4-2q^4=r^2$

$p = 3, 113, 57123, 262621633,\dots$ (even entries of A360187, Generalized Somos-5 sequence)

$q = 2, \,84,\, 6214,\,151245528,\dots$ (no sequence yet?)

Answer 1

$2a^4+b^4=2d^2$ where we can assume $a,b,d$ are pairwise coprime.

Let $b=2x$, then $(d-a^2)(d+a^2)=8x^4$. The g.c.d. of $d-a^2$ and $d+a^2$ is $2$ and so these two numbers are $2p^4$ and $4q^4$ (or minus these numbers) where $x=pq$.

Then $a^2=|p^4-2q^4|, d^2=(p^4+2q^4)^2$ and $b=2pq$, as required.