On the equation $a^4+b^4+c^4=2d^4$ in natural numbers with $a<b<c<d$

Question

I asked a simillar question with the weaker restriction:

On the equation $a^4+b^4+c^4=2d^4$ in positive integers $a\lt b\lt c$ such that $a+b\ne c$

.

---

I couldn't find any solution to this equation. And if $a^2+2ab+b^2=c^2$, then $c>d=a+b$.

Main question: Find some solutions to the equation $a^4+b^4+c^4=2d^4$ in natural numbers with $a<b<c<d$.

Thanks for advance.

Answer 1

If we take a point $3Q(m)$ then we get a solution as follows.
For more details, please see $a^4+b^4+c^4=2d^4$.

      a =  100954906225546184690686373445232988785377384105455647295649619
           847550243590542235300309485123275004949554849358738273723252717
           90197500095135824990543003573711100267705489

      b =  182720084009424134346291581847691617723750250933518665717300613
           005234908926888631952883207666518927841277103892927263065478117
           85836937759879470109480968700349292383907888

      c =  186724255151010815056751706873402208434522449763207787663872587
           945588542980120170792502069164146307065641765086510884090297462
           19862671046394577805380183600483828969086905

      d =  186780052966035701609386026218590575760779098084873786620653552
           576571949525574948356153636564402836837404166574064647539252512
           98760140535473903764903746388941211178351357