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 $$(1,1,2,3),(7,7,12,113),(7967,7967,18984,262621633),...$$

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Edit (by Piezas, Jan 10, 2025):

Just to emphasize that the relations given by the OP to $2x^4-y^4=z^2$ are not unique. Given,

$$ax^4-y^4=z^2$$

then $(x,y,z)$ also solves both,

$$z^4 + 4a(x y)^4 = (a x^4 + y^4)^2$$ $$(a x^4 + y^4)^4 - a(2x y z)^4 = (4a x^4 y^4 - z^4)^2$$

Also, note that if there is one solution to $p^4-aq^4=r^2$, then another is,

$$(p^4 + a q^4)^4 - a(2p q r)^4 = (4a p^4q^4 - r^4)^2$$

though the numbers get large very fast. For the case $a=2$, if,

$$p^4-2q^4=r^2$$

then it leads to one of the OP's questions,

$$r^4 + r^4 + (2p q)^4 = 2(p^4 + 2 q^4)^2$$

proving infinitely many solutions to $x^4+x^4+y^4=2z^2$ with $\text{gcd}(x,y)=1$. Using the recursion and starting with,

$$1^4+1^4+2^4 = 2\times3^2$$

Since $2\times1\times2\times3=\color{blue}{12}$ then,

$$7^4+7^4 +\color{blue}{12}^4 = 2\times113^2$$

Since $2\times7\times12\times113=\color{blue}{18984}$ then,

$$7967^4+7967^4+\color{blue}{18984}^4 =2\times262621633^2$$

Since $2\times7967\times 18984\times262621633 = \color{blue}{79440695094614448}$ then,

$$60912456065182847^4 + 60912456065182847^4 + \color{blue}{79440695094614448}^4 = 2z^2$$

and so on, with the numbers getting larger as expected, though it may miss smaller solutions like $26^4 + 239^4 + 239^4 = 2\times57123^2$ pointed out by Empy2 in the comments.

Answer 2

(This answer was moved from the related post "Diophantine equation $a^4+b^4+c^4=2d^2$" as it seems more suited here.)

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I. General case

We generalize Empy2's answer from that post and show that for any,

$$x^4+y^4+z^4 = 2w^2$$

then it can be used as a "seed" to create a polynomial parameterization. Employing Empy2's form,

$$(x(n^2 + b) + c n)^4 + (y(n^2 + b) + d n)^4 + (z(n^2 - b))^4 = 2(w(n^2 + b)^2 + p(n^2 + b)n + q n^2)^2$$

(with minor changes in the variables for aesthetics) expand and collect powers of $n$. This is an octic polynomial in $n$ that turns out to be semi-palindromic,

$$c_0 n^8+ c_1 n^7+ c_2 n^6+ c_3 n^5+ c_4 n^4+ c_3 b n^3+ c_2 n^2 n^2+ c_1 b^3 n + c_0 b^4 =0$$

where the $c_k$ are polynomials in the unknowns. Thus, there are only five equations $c_0 = c_1 = c_2 = c_3 = c_4 =0$ to be solved which can be done. Ignoring numerical factors, then,

\begin{align} c_0 &:= \color{red}{(x^4+y^4+z^4-2w^2)} = 0\\ c_1 &:= \color{blue}{(p w - c x^3 - d y^3)} = 0\\ c_2 &:= \color{green}{(p^2 + 2 q w - 3 c^2 x^2 - 3 d^2 y^2)} - 2 b (x^4 + y^4 - z^4 - 2 w^2) = 0\\ c_3 &:= (p q - c^3 x - d^3 y) + 3 b \color{blue}{(p w - c x^3 - d y^3)} = 0\\ c_4' &:= c^4 + d^4 - 2 q^2 - 4 b \color{green}{(p^2 + 2 q w - 3 c^2 x^2 - 3 d^2 y^2)} + 6 b^2\color{red}{(x^4+y^4+z^4-2w^2)} = 0\\ c_4 &:= c^4 + d^4 - 2 q^2 - 4 b \color{green}{(p^2 + 2 q w - 3 c^2 x^2 - 3 d^2 y^2)}=0 \end{align}

As one can see, the system has a lot of symmetries which help in solving it plus $c_4$ is just linear in $b$ (since the red part vanishes). Solve the easiest parts first, or $c_1 = c_3 = 0$ for $(p,q)$, then substitute into $c_2 = c_4 = 0$. They are only linear in $b$, so eliminate it using resultants. As pointed out by Empy2, one ends up with a sextic polynomial in $(c,d)$.

However, it only has even powers of $w$. Substituting $w = \sqrt{\dfrac{x^4+y^4+z^4}2}$ into the sextic, it conveniently factors into three quadratics. (Why?) Two quadratics yield only complex roots, but the third one (factored over a square root) is,

$$\frac{c\,m}{d\,m} = \frac{x\,y\,(x^4+y^4+z^4-2x^2y^2)\mp(x^2-y^2)z^2\sqrt{2(x^4+y^4+z^4)}}{y^2(x^4+y^4+z^4)-2x^2(y^4+z^4)}$$

where $m$ is just a scaling factor so the other variables below that depend on $(c,d)$ will be integers. Thus, if one has any solution to $x^4+y^4+z^4 = 2w^2$, then the expression under the square root is a square and we get rational $c/d$. The five unknowns $(b,c,d,p,q)$ of Empy2's form are then,

\begin{align} b\, &= \frac{-(p^2+2qw-3c^2x^2-3d^2y^2)}{4z^4}\\ p\, &= \frac{cx^3+dy^3}{w}\\ q\, &= \frac{(c^3x+d^3y)\,w}{cx^3+dy^3\;}\\ \frac{c}{d} &= \;\text{expression above} \end{align}

plus any solution to $x^4+y^4+z^4=2w^2$.

Note: Similarly, any solution to $a^3+b^3+c^3=d^3$ and $a^4+b^4+(a+b)^4+c^4+d^4 = e^4$ yields a polynomial parameterization, so $a^4+b^4+c^4=2d^2$ joins the club. (See also this post.)

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II. Examples

Since we can permute the $4$th powers without affecting $w$, then there are many values to $c/d$ for a given $\pm(x,y,z,w)$. For example, using the negative case $\mp$ of the formula for $c/d$,

\begin{align} \frac{c}{d};\;\, & (x,y,z)\qquad\qquad\\[4pt] \frac{2\times17}{2\times7}=\frac{\color{blue}{34}}{14};\; & (3,7,14)\\[4pt] \frac{7\times167}{7\times283}=\frac{\color{blue}{1169}}{1981};\; & (7,14,3)\\[4pt] \frac{1\times2723}{1\times1567}=\frac{\color{blue}{2723}}{1567};\; & (14,3,7)\end{align}

and we get,

\begin{align} (3(n^2 - 6) + \color{blue}{34} n)^4 + (7(n^2 - 6) + 14 n)^4 + (14(n^2 + 6))^4 = 2w_1^2\\ (7(n^2 + 4350) + \color{blue}{1169} n)^4 + (14(n^2 + 4350) + 1981 n)^4 + (3(n^2 - 4350))^4 = 2w_2^2\\ (14(n^2 + 6630) + \color{blue}{2723} n)^4 + (3(n^2 + 6630) + 1567 n)^4 + (7(n^2 - 6630))^4 = 2w_3^2\end{align}

and so on. More can be found by varying the sign of $\pm(x,y,z,w)$ and the positive case $\mp$ of the formula for $c/d$. When signed, these polynomials are $x+y+z\neq0$ as requested by the OP.

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III. Special case $x=y$

As pointed out by Empy2, when $x=y$ in the general case, then $c/d=-1$. This implies $p=0$ but there is division by zero for $q$. So one has to derive the solution in a different manner. Adjusting our variables for aesthetics, given,

$$x^4+x^4+y^4=2(\pm w)^2$$

then,

$$x^4(n^2 + y n - b)^4 + x^4(n^2 - y n - b)^4 + y^4(n^2 + b)^4 = 2(w(n^2 - b)^2 + c y^2 n^2)^2$$

where,

$$b = \frac14(4w+3y^2),\quad c=3w+2y^2$$

and the $\pm w$ implies identities come in pairs. For example, given the OP's

$$7^4+7^4+12^4=2(\pm113)^2$$

then $w=-113$ and $w=113$ yields,

$$7^4(n^2 + 12 n + 5)^4 + 7^4(n^2 - 12 n + 5)^4 + 12^4(n^2 - 5)^4 = 2w_1^2$$

$$7^4(n^2 + 12 n - 221)^4 + 7^4(n^2 - 12 n - 221)^4 + 12^4(n^2 + 221)^4 = 2w_2^2$$

respectively. And so on.

Answer 3

$$ a^4+b^4+c^4=2d^2 \tag{1} $$ Let $x=a, v=2d.$ Then we get the quartic equation. $$ v^2 = 2x^4+2b^4+2c^4$$

For instance, we consider the case for $(b,c)=(2,1).$

This quartic equation is birationally equivalent to an elliptic curve below. $$E: Y^2 = X^3-17X$$
with $$x = (-17X-17-3Y)/(-3Y+X+17) $$ $$v = (544Y+306X^2-1734+54X^3+918X)/((-3Y+X+17)^2) $$

$E$ has rank $2$ and generators are $(X,Y)=(-1,-4),(9,-24).$ Since $E$ has infinitely many rational solutions, equation $(1)$ has infinitely many integer solutions. By using group structure, we get the solutions. Here are several small solutions below.

 [a,b,c,d]
[45, 14, 7, 1439]
[55, 2, 1, 2139]
[121, 46, 23, 10467]
[761, 766, 383, 592107]
[387575, 229682, 114841, 112962961059]
[713085, 169826, 84913, 360170851159]
[5184425807, 2354091262, 1177045631, 19430283429551362323]
[23049805967, 80860770142, 40430385071, 4780468378946494597443]

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The special case $c=b:$

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

$$E: Y^2 = X^3-2X$$
$E$ has rank $1$ and generator is $(X,Y)=(-1,-1).$

Here are several small solutions below.

[a,b,d]
[2, 1, 3]
[12, 7, 113]
[26, 239, 57123]
[18984, 7967, 262621633]
[4096150, 2750257, 14070212996451]
[709924644, 3262580153, 10650393355715621873]

Answer 4

Here is an infinite number of nontrivial solutions as requested by OP in (v).
There are simple ways to tweak them, for example shift the common factor $2$ shared by the coefficients of $n^2$, to the constants, so the first term becomes $(3n^2-2n-2)$ and so on.
I have found several similar quadratic solutions by bruteforce, but they all seem to boil down to variations on these two.

In particular, take the nine coefficients from the three quadratics involved, and form a $3×3$ matrix. In all quadratic solutions I have found, the determinant always has the form $96k^3$ or $1020k^3$.

$$(6n^2-2n-1)^4+(6n^2+2n-1)^4+(12n^2+2)^4=2(108n^4+32n^2+3)^2,\\ (6n^2-5)^4+(6n^2+n+5)^4+(12n^2+19n+10)^4=2(108n^4+306n^3+449n^2+255n+75)^2$$