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 $a=n$ and six known solutions are,

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

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

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

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

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

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

$b_7 = ??$

where $b_6$ was found by Seiji Tomita.

Question:

Being an elliptic curve, there are infinitely many $b_m$. 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. Note that 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.