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

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  • $\begingroup$ The equation $$a^8+b^8+c^8=2d^2$$ is derived from: $$\big\{p(x^2-y^2)^4+q\big\}^2+\big\{p(2xy)^4+q\big\}^2+\big\{p(x^2+y^2)^4+q\big\}^2=2\big\{p(x^8+14x^4y^4+y^8)+q\big\}^2+q^2$$ where $(p,q)=(1,0) $ so I don't believe there are alternative solutions to your first question that don't satisfy the constraint. $\endgroup$ – Mr Pie Jan 21 at 12:27
  • $\begingroup$ It appears that for $n>2$ there is no solution. With which search limit would you be content ? $\endgroup$ – Peter Jan 21 at 18:29
  • $\begingroup$ For $n>2$, I guess good evidence would be no solution for $a \leqslant b \leqslant c \leqslant 100000$ for $n \leqslant 100$. I wouldn't mind $c \leqslant 10000$ either, but the former is preferable as an answer. For $n=2$, I would be satisfied with no non-Pythagorean solutions for $a \leqslant b \leqslant c \leqslant 1000000$. But I guess that would be almost impossible, so $c \leqslant 100000$ would be okay, but again, the former is preferable as an answer. $\endgroup$ – Haran Jan 22 at 11:46
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    $\begingroup$ Applying standard equal sums of squares parameterizations (cf. Bradley et al.), there exist integers $m,j,r,s,t$ such that, in the general case, \begin{align} d &= \frac{(mr-js)(r^2+s^2+t^2)}{2(r+s)}, \\ a^{2n} &= t(js-mr), \\ b^{2n} &= \frac{(mr-js)\bigl(t^2-\bigl((r+s)^2-2s^2\bigr)\bigr)}{2(r+s)}, \\ c^{2n} &= \frac{(mr-js)\bigl(t^2+\bigl((r-s)^2-2s^2\bigr)\bigr)}{2(r+s)}, \end{align} or some other permutation of $a,b,c$ (though there may be an easy 'WLOG' argument?). Perhaps you can use this, with your constraint $\gcd(a,b,c)=1$, to move forward? $\endgroup$ – Kieren MacMillan Jan 22 at 15:55
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    $\begingroup$ select(r->r[1]%9==0 || r[1]%9==2 || r[1]%9==5 || r[1]%9==8,%) %3 = [[0, 0, [0, 0]], [2, 0, [1, 1]], [2, 1, [0, 1]], [5, 0, [1, 4]], [5, 1, [0, 4]], [5, 4, [0, 1]], [8, 0, [1, 7]], [8, 0, [4, 4]], [8, 1, [0, 7]], [8, 4, [0, 4]], [8, 7, [0, 1]], [9, 1, [1, 7]], [9, 1, [4, 4]], [9, 4, [1, 4]], [9, 7, [1, 1]], [11, 0, [4, 7]], [11, 4, [0, 7]], [11, 7, [0, 4]], [14, 0, [7, 7]], [14, 7, [0, 7]], [18, 4, [7, 7]], [18, 7, [4, 7]]] $\endgroup$ – user645636 Jan 28 at 16:08
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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 $$(1,1,2,3),(7,7,12,113),(7967,7967,18984,262621633),...$$

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