To solve,

$$A^4+B^4 = C^4+D^4$$

we use Euler's method. Let,

$$(p+q)^4+(r-s)^4=(p-q)^4+(r+s)^4$$

and define $p = (a^3 - b),\, q = a y,\, r = b (a^3 - b),\, s = y.\,$ The equation above transforms to the simple form,

$$(a^3 - b) (b^3 - a) = y^2$$

This is birationally equivalent to an elliptic curve. Assume the case $a=n.$ Six "smallish" solutions are known (with $b_6$ found by Seiji Tomita):

$b_1 =\frac{n\,(\color{red}{1} + n)}{(1 + n)}$

$b_2 =\frac{n\,(\color{red}{4} + n^2 + 10n^4 + n^6)}{(1 + 10n^2 + n^4 + 4n^6)}$

$b_3 =\frac{n\,(\color{red}{9} - 44n^2 + 190n^4 + 100n^6 + n^8)}{(1 + 100n^2 + 190n^4 - 44n^6 + 9n^8)}$

$b_4 =\frac{n\,(\color{red}{16} - 543n^2 + 4632n^4 + 15100n^6 + 10632n^8 + 22758n^{10} + 6568n^{12} + 5820n^{14} + 552n^{16} + n^{18})}{(1 + 552n^2 + 5820n^4 + 6568n^6 + 22758n^8 + 10632n^{10} + 15100n^{12} + 4632n^{14} - 543n^{16} + 16n^{18})}$

$b_5 =\frac{n\,(\color{red}{25} -3524n^2 + 113482n^4 + 979388n^6 + 1486687n^8 + 2379064n^{10} + 5807660n^{12} + 3492760n^{14} + 2404327n^{16} + 45836n^{18} + 69418n^{20} +2092n^{22} + n^{24})\quad}{(1 + 2092n^2 + 69418n^4 + 45836n^6 + 2404327n^8 + 3492760n^{10} + 5807660n^{12} + 2379064n^{14} + 1486687n^{16} + 979388n^{18} + 113482n^{20} - 3524n^{22} + 25n^{24})}$

$b_6 = \frac{(n^{38}+6234n^{36}+569433n^{34}-1574764n^{32}+165024372n^{30}+965109816n^{28}+4050441732n^{26}+8937136896n^{24}+11838786414n^{22}+16534395580n^{20}+11971009518n^{18}+9409389288n^{16}+3853491204n^{14}+973803384n^{12}-132081036n^{10}+119823968n^8+32622105n^6+1538106n^4-15551n^2+\color{red}{36})n}{(36n^{38}-15551n^{36}+1538106n^{34}+32622105n^{32}+119823968n^{30}-132081036n^{28}+973803384n^{26}+3853491204n^{24}+9409389288n^{22}+11971009518n^{20}+16534395580n^{18}+11838786414n^{16}+8937136896n^{14}+4050441732n^{12}+965109816n^{10}+165024372n^8-1574764n^6+569433n^4+6234n^2+1)}$

$b_7 = \;?$

$b_8 = \;?$

Update 1. As pointed out by Sidharth Ghoshal:

1. The coefficients of $b_m$ sum to $2^k$, namely $2^1,\,2^4,\,2^8,\,2^{16},\,2^{24},\,2^{36}.$
2. The degree $d$ of the denominators are $1, 6, 8, 18, 24, 38$.
3. So he pointed out it seems both the power $k$ and degree $d$ are functions of $m$.

$$\begin{array}{|c|c|c|c|} \hline m&m^2&k&d\\ \hline\color{blue} 1&1&1&1\\ \hline 2&4&4&6\\ \hline\color{blue} 3&9&8&8\\ \hline 4&16&16&18\\ \hline\color{blue} 5&25&24&24\\ \hline 6&36&36&38\\ \hline\color{blue} 7&49&?&?\\ \hline 8&64&?&?\\ \hline\color{blue} 9&81&80&80\\ \hline \end{array}$$

Hopefully someone can find $b_7$ and $b_8$ so we can test this "pattern".

Update 2. I found another solution, not so smallish. I don't know if this is really $b_9$, but it fits the pattern above.

$b_9 \overset{?}= \frac{n(\color{red}{81} - 407672n^{2} + 482840284n^{4} + 77282464024n^{6} - 336916143634n^{8} - 269073613222472n^{10} + 1757845056647068n^{12} + 96071436547023656n^{14} + 1804737030046873645n^{16} + 14686441538438168608n^{18} + 81169043718961628976n^{20} + 234981869722959440480n^{22} + 521545149005486710568n^{24} + 1531812067426248122976n^{26} + 4989911070816764106352n^{28} + 16007166226577445125920n^{30} + 37939061001155279992178n^{32} + 74373521995862636237296n^{34} + 120310095858827397125128n^{36} + 161721875649245818566864n^{38} + 188963554058602494368852n^{40} + 186242485219018870892816n^{42} + 160579318360111593624264n^{44} + 117574894628368241794864n^{46} + 74255179165990324638802n^{48} + 38818758949934604483488n^{50} + 17178808479864835237360n^{52} + 5723686723196286338272n^{54} + 1516563414177942983208n^{56} + 299652084748204903904n^{58} + 42614192414467088816n^{60} + 2343223081001783712n^{62} + 159626691817966525n^{64} + 68841496974471880n^{66} + 3884690979741692n^{68} + 88440504999640n^{70} + 2931242386030n^{72} - 7601467144n^{74} + 67097980n^{76} + 70888n^{78} + n^{80})}{1 + 70888n^{2} + 67097980n^{4} - 7601467144n^{6} + 2931242386030n^{8} + 88440504999640n^{10} + 3884690979741692n^{12} + 68841496974471880n^{14} + 159626691817966525n^{16} + 2343223081001783712n^{18} + 42614192414467088816n^{20} + 299652084748204903904n^{22} + 1516563414177942983208n^{24} + 5723686723196286338272n^{26} + 17178808479864835237360n^{28} + 38818758949934604483488n^{30} + 74255179165990324638802n^{32} + 117574894628368241794864n^{34} + 160579318360111593624264n^{36} + 186242485219018870892816n^{38} + 188963554058602494368852n^{40} + 161721875649245818566864n^{42} + 120310095858827397125128n^{44} + 74373521995862636237296n^{46} + 37939061001155279992178n^{48} + 16007166226577445125920n^{50} + 4989911070816764106352n^{52} + 1531812067426248122976n^{54} + 521545149005486710568n^{56} + 234981869722959440480n^{58} + 81169043718961628976n^{60} + 14686441538438168608n^{62} + 1804737030046873645n^{64} + 96071436547023656n^{66} + 1757845056647068n^{68} - 269073613222472n^{70} - 336916143634n^{72} + 77282464024n^{74} + 482840284n^{76} - 407672n^{78} + 81n^{80}}$

Question:

Given this elliptic curve, from an initial rational point $b_m$, we know there are infinitely more. Can we predict that $b_7$'s numerator has a linear term that is $\color{red}{49}$? Or for general $b_m$, is it $m^2$?

P.S. Another curious feature is the coefficients of the numerator and denominator are palindromic wrt to each other. (A related question was asked in this post but focuses on another aspect.)