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 $a=n.$ <strike>*Six*</strike> *Seven* "smallish" solutions are known (with $b_6$ and $b_7$ found by Seiji Tomita):
$b_1 =\frac{n\,(\color{red}{1} + n)}{(1 + n)}$
$b_2 =\frac{n\,(\color{red}{4} + n^2 + 10n^4 + n^6)}{(1 + 10n^2 + n^4 + 4n^6)}$
$b_3 =\frac{n\,(\color{red}{9} - 44n^2 + 190n^4 + 100n^6 + n^8)}{(1 + 100n^2 + 190n^4 - 44n^6 + 9n^8)}$
$b_4 =\frac{n\,(\color{red}{16} - 543n^2 + 4632n^4 + 15100n^6 + 10632n^8 + 22758n^{10} + 6568n^{12} + 5820n^{14} + 552n^{16} + n^{18})}{(1 + 552n^2 + 5820n^4 + 6568n^6 + 22758n^8 + 10632n^{10} + 15100n^{12} + 4632n^{14} - 543n^{16} + 16n^{18})}$
$b_5 =\frac{n\,(\color{red}{25} -3524n^2 + 113482n^4 + 979388n^6 + 1486687n^8 + 2379064n^{10} + 5807660n^{12} + 3492760n^{14} + 2404327n^{16} + 45836n^{18} + 69418n^{20} +2092n^{22} + n^{24})\quad}{(1 + 2092n^2 + 69418n^4 + 45836n^6 + 2404327n^8 + 3492760n^{10} + 5807660n^{12} + 2379064n^{14} + 1486687n^{16} + 979388n^{18} + 113482n^{20} - 3524n^{22} + 25n^{24})}$
$b_6 = \frac{n(n^{38}+6234n^{36}+569433n^{34}-1574764n^{32}+165024372n^{30}+965109816n^{28}+4050441732n^{26}+8937136896n^{24}+11838786414n^{22}+16534395580n^{20}+11971009518n^{18}+9409389288n^{16}+3853491204n^{14}+973803384n^{12}-132081036n^{10}+119823968n^8+32622105n^6+1538106n^4-15551n^2+\color{red}{36})}{(36n^{38}-15551n^{36}+1538106n^{34}+32622105n^{32}+119823968n^{30}-132081036n^{28}+973803384n^{26}+3853491204n^{24}+9409389288n^{22}+11971009518n^{20}+16534395580n^{18}+11838786414n^{16}+8937136896n^{14}+4050441732n^{12}+965109816n^{10}+165024372n^8-1574764n^6+569433n^4+6234n^2+1)}$
$b_7 = \frac{n(n^{48}+15704n^{46}+3430692n^{44}-57632376n^{42}+6702252562n^{40}+75079777160n^{38}+707080531380n^{36}+3414184912920n^{34}+7188385885663n^{32}+19104201262320n^{30}+33722429304776n^{28}+46168990682064n^{26}+57120264919228n^{24}+47015521855632n^{22}+37697060130056n^{20}+19106431585968n^{18}+8759831502031n^{16}+1465420875576n^{14}-8798478828n^{12}-70938314968n^{10}+2565235282n^8+603165288n^6+13866564n^4-54088n^2+\color{red}{49})}{(49n^{48}-54088n^{46}+13866564n^{44}+603165288n^{42}+2565235282n^{40}-70938314968n^{38}-8798478828n^{36}+1465420875576n^{34}+8759831502031n^{32}+19106431585968n^{30}+37697060130056n^{28}+47015521855632n^{26}+57120264919228n^{24}+46168990682064n^{22}+33722429304776n^{20}+19104201262320n^{18}+7188385885663n^{16}+3414184912920n^{14}+707080531380n^{12}+75079777160n^{10}+6702252562n^8-57632376n^6+3430692n^4+15704n^2+1)}$
$b_8 = \;?$
---
**Update 1.** As pointed out by Sidharth Ghoshal:
1. The coefficients of $b_m$ sum to $2^k$, namely $2^1,\,2^4,\,2^8,\,2^{16},\,2^{24},\,2^{36},\,2^{48}.$
2. The degree $d$ of the denominators are $1, 6, 8, 18, 24, 38, 48$.
3. So he pointed out it seems both the power $k$ and degree $d$ are functions of $m$.
$$\begin{array}{|c|c|c|c|}
\hline
m&m^2&k&d\\
\hline\color{blue}
1&1&1&1\\
\hline
2&4&4&6\\
\hline\color{blue}
3&9&8&8\\
\hline
4&16&16&18\\
\hline\color{blue}
5&25&24&24\\
\hline
6&36&36&38\\
\hline\color{blue}
7&49&48&48\\
\hline
8&64&?&?\\
\hline\color{blue}
9&81&80&80\\
\hline
\end{array}$$
Hopefully someone can find $b_8$ so we can close this table.
**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 $b_8$'s numerator has a linear term that is $\color{red}{64}$? 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][1] but focuses on another aspect.)
[1]: https://math.stackexchange.com/questions/4779869/finding-rational-points-on-the-elliptic-curve-for-a4b4-c4d4