**Background:** The equation


$$a^4+b^4+c^4=2d^4$$


has infinitely many positive integral solutions if we take $c=a+b$ and $a^2+ab+b^2=d^2$ further assuming that $GCD(a,b,c)=1$.


**Main problem:** Find some positive integral solutions to the equation


$$a^4+b^4+c^4=2d^4$$


with $a\lt b\lt c\ne a+b$ and $GCD(a,b,c)=1$.



On the equation $a^4+b^4+c^4=2d^4$ in coprime positive integers $a\lt b\lt c$ such that $a+b\ne c$
Let $A=\frac{a}{d}, B=\frac{b}{d}, C=\frac{c}{d},$ then we get


$$A^4+B^4+C^4=2\tag{1}$$


Let $A=x+y, B=x-y, z=C^2$ then


$$2x^4+12y^2x^2+2y^4+z^2 = 2\tag{2}$$


Hence


$$y^2 = -3x^2 \pm \frac{\sqrt{32x^4+4-2z^2}}{2}\tag{3}$$


So we find the rational solutions of $(4)$.


$$v^2=32x^4+4-2z^2\tag{4}$$


$(4)$ can be parameterized to $(5)$ using $(z,v)=(4x^2,2)$ with $w=(k^2+2)C$.


$$w^2 = 4(k^2+2)(-2+k^2)x^2-4(k^2+2)k\tag{5}$$


Thus we must find the rational solutions of simultaneous equations $(3),(5)$.


First we get a parametric solution of $(5)$ using giving some rational number $k$.


Take $k=\frac{-9}{13}$. Then we get:


$$x = \frac{3}{142}\frac{28561m^2-430732-7081100m}{28561m^2+430732}$$
$$w = \frac{-1257}{11999}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$
$$C = \frac{-3}{71}\frac{-10768300+173732m+714025m^2}{28561m^2+430732}$$


Hence $(3)$ becomes


$$y^2 = \frac{10083247442281m^4-41255619857608m^2+2788930240200m^3+2293337020040464-42060204482400m}{20164(28561m^2+430732)^2}$$


Hence we must find the rational solutions $(m,V)$ of $(6)$  

$$V^2 = 10083247442281m^4+2788930240200m^3-41255619857608m^2-42060204482400m+2293337020040464\tag{6}$$


Quartic equation $(6)$ is birationally equivalent to the elliptic curve using a known solution $(m,V)=(360/169,-47445892)$ as follows.
$$Y^2 = X^3 -X^2 -1097465452X+ 3288951361780\tag{7}$$


Though I couldn't find the generator of elliptic curve , I got one solution
$P(X,Y)=(\cfrac{-97636631990}{5536609}, \cfrac{53963430434179560}{13027640977})$.


Since $(7)$ has rank $1$, we get infinitely many solutions of $(7)$ using group law as follows.


For instance, we get a solution of $(6)$,  

$2Q(m,V)=(\cfrac{-762617488871059540}{36474714629699307},\cfrac{-63980049963485293932709632365019549028}{46581170383319934672021751209})$


We get $(x,y)=(\cfrac{4317501713916820330332746705607}{16486256234272723829845043984402},\cfrac{12346241358503326020929860043561}{16486256234272723829845043984402})$


Finally, we get


$a = 4014369822293252845298556668977$  

$b = 8028813922964684804294250901215$  

$c = 8331871536210073175631303374584$  

$d = 8243128117136361914922521992201$


