The equation $a^{4n}+b^{4n}+c^{4n}=2d^2$

Question

Recently, I found that if $a+b=c$, then $a^4+b^4+c^4=2d^2$ for some positive integer $d$. The parametric equation is: $$m^4+n^4+(m+n)^4=2(m^2+mn+n^2)^2$$ The condition $a+b=c$ (assuming $c \geqslant a,b$) isn't necessary. For example: $$7^4+7^4+12^4=2 \cdot 113^2$$ We can note that when we make the equation in the form $a^{4n}+b^{4n}+c^{4n}=2d^2$, and we impose the condition $a^n+b^n=c^n$ for the parametric solution:

(i) When $n=1$, we can have any positive integers $a+b=c$

(ii) When $n=2$, we can have any Pythagorean Triple $(a,b,c)$.

(iii) When $n>2$, there are no solutions by Fermat's Last Theorem.

Checking when $n=2$, I saw that there are no solutions for $a \leqslant b \leqslant c \leqslant 3000$ where $a^2+b^2 \neq c^2$. I have not run a program for any value $n>2$ though.

For positive integers $a \leqslant b \leqslant c$ where $\gcd(a,b,c)=1$ :

$1$. Are there any solutions for $a^8+b^8+c^8=2d^2$ where $a^2+b^2 \neq c^2$ ?

$2$. Are there any solutions for $a^{4n}+b^{4n}+c^{4n}=2d^2$ where $n>2$?

$3$. For the solutions of $a^4+b^4+c^4=2d^2$ which do not follow $a+b=c$, is there any way of generating more solutions from primitive solutions? From primitive solution $(a,b,c,d)$, can we get more solutions $(A,B,C,D)$?

EDIT : First off, it suffices to focus on solutions for $a^{4n}+b^{4n}+c^{4n}=2d^2$ for prime $n$ alone, since if we have a solution for some $n$, then we have a solution for the divisors of $n$ as well. An accepted answer would be one of:

$(i)$ Verifying problem $1$ for $a \leqslant b \leqslant c \leqslant 1000000$.

$(ii)$ Verifying problem $2$ for $a \leqslant b \leqslant c \leqslant 100000$ (for odd primes $n<100$).

$(iii)$ Verifying problem $1$ for $a \leqslant b \leqslant c \leqslant 100000$ and problem $2$ for $a \leqslant b \leqslant c \leqslant 10000$ (for odd primes $n<100$).

$(iv)$ Proof or Counterexample for either problems $1$ or $2$.

$(v)$ Relations, generation or parametric characterization of the non-trivial solutions of $$a^4+b^4+c^4=2d^2$$

Answer 1

Problem 3

This is a scheme to generate the solutions which, like your example of $(7,7,12,113)$, have two of $a,b,c$ equal.

Consider the following system of three closely related equations.

E: $2x^4-y^4=z^2$

F: $x^4+8y^4=z^2$

G: $x^4-2y^4=z^2$

A 'base solution' $(x,y,z)$ of E can be used to generate a solution $(z,xy,2x^4+y^4)$ of F.

Each solution $(x,y,z)$ of F can be used to generate a solution $(z,2xy,|x^4-8y^4|)$ of G.

Each solution $(x,y,z)$ of G can be used to generate a further solution $(z,xy,x^4+2y^4)$ of F.

Each solution $(x,y,z)$ of F can be used to generate the solution $(x,x,2y,z)$ of the required equation.

Example starting with the solution $(1,1,1)$ of E.

The scheme generates F$(1,1,3)$, G$(3,2,7)$,F$(7,6,113)$, G$(113,84,7967)$, F$(7967,9492,262621633)$, .....

The required solutions are then $$(1,1,2,3),(7,7,12,113),(7967,7967,18984,262621633),...$$