$a^4+b^4+c^4=2d^2$ such that $a,b,c,d$ are all nonzero Integers & $a+b+c \ne0$.

Question

Let $a^4+b^4+c^4=2d^2$ such that $a,b,c,d$ are all nonzero Integers & $a+b+c \ne0$.

Already I know, Jacobi - Maiden Identity and it is Generalized by Ramanujan :

$$a^4 + b^4 + c^4 = 2(ab+ac+bc)^2$$ Where $a+b+c=0$

I Find four Solutions by Trial & Error $$(a,b,c,d)=(1,3,10,71),(1,9,22,347),(1,12,41,1193),(2,55,111,8971)$$

knowing one solution I will generate Infinitely many solutions by using Elliptic curves :

Consider $a^4+b^4+c^4=2d^2$ ---(1)

Wkt $1^4+3^4+10^4=2×71^2$

Let $a=x+3$, $b=x+10$, $c=1$ and $d=y$

$(x+3)^4+(x+10)^4+1^4=2(x^4+26x^3+327x^2+2054x+71^2)$

$x^4+26x^3+327x^2+2054x+5041=y^2$

We need to find Rational solutions for above Equation.

Since it has non trivial rational point so Above Quartic Equation is birationally equivalent to an Elliptic curve $E$ below

$E$ : $V^2=U^3-194643U-71422$

$$x=\frac{V-39U-17199}{6U+2646}$$

$$y=\frac{18x^2+234x-U+981}{18}$$

Rank of an Elliptic curve $E$ is 2

Generators:$(-297,5616)$,$(-9,1296)$

Since Rank of $E$ is 2 so (1) has Infinitely many Integer Solutions with $a+b+c \ne0$

I find Some Solutions by using Group structure of an Elliptic curve $E$ :

$$(a,b,c,d) = (3,7,14,143), (9,14,89,5603), (13,105,196,28261),(10,173,243,46811), (923,1596,4865,16843381)$$

Is there any parametrization that gives Infinitely many solutions ?

Is there any Complete Solution ?

Answer 1

It can be shown there are infinitely many polynomial identities to $x^4+y^4+z^4=2t^2$ such that $\color{red}{x+y+z\neq0}$ and $\text{gcd}(x,y,z)=1.$

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I. Family 1

$$2^4(an^2+b)^4 + (an^2 - 2mn - b)^4 + (an^2 + 2mn - b)^4 = 2(3a^2n^4 + 32m^2n^2+3b^2)^2$$

where $b=6m^2/\color{red}a\,$ for any $(m,a)$. For example, let $m=1$, thus $b=6/a$, but we choose $a$ such that $b$ is also an integer.

Let $a=6$, $b=1$, so,

$$2^4(6n^2+1)^4+(6n^2−2n−1)^4+(6n^2+2n−1)^4=2(108n^4+32n^2+3)^2$$

Let $a=3$, $b=2$, so,

$$2^4(3n^2+2)^4+(3n^2−2n−2)^4+(3n^2+2n−2)^4=2(27n^4+32n^2+12)^2$$

where the first is Empy2's identity in the comments, and so on for any $(m,a)$.

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II. Family 2

$$(an^2-b)^4 + ((an^2 + b)m + cn)^4 + ((an^2 + b)(m+1) + dn)^4 = 2t^2$$

where,

\begin{align} b &= (2m^2 + 2m + 1)(m^2 + m + 1)(m + 1)m/\color{red}a\\[4pt] c &= 2m^4 + 2m^3 - 2m - 1\\[4pt] d &= 2m^4 + 6m^3 + 6m^2 + 4m + 1 \end{align}

for any $(m,a)$. As usual, we choose $a$ such that $b$ is also an integer. For example, let $m=1$, then $b=30/a$, and we have many integer choices for $a$.

Let $a=6$, $b=5$, so,

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

Let $a=3$, $b=10$, so,

$$\;(3n^2-10)^4+(3n^2+10+n)^4+(2(3n^2+10)+19n)^4 =\\ 2(27n^4 + 153n^3 + 449n^2 + 510n + 300)^2$$

and so on for any $(m,a)$.

Update: Empy2 suggested a more symmetrical version,

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

where,

\begin{align} b &= (x^2 + y^2)(x^2 + x y + y^2)x y/\color{red}a\\[4pt] c &= 2x^3y - (x + y)^3(x - y)\\[4pt] d &= 2xy^3 + (x + y)^3(x - y) \end{align}

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III. Derivation

To find such identities, what you do is assume three quadratic polynomials $P_k(x)$ and one quartic $Q_1(x)$, then equate,

$$P_1(x)^4+P_2(x)^4+P_3(x)^4 = 2Q_1(x)^2$$

and collect powers of $x$. Set all coefficients equal to zero, and you have a system of $m$ equations in $n$ unknowns, which is near-impossible to solve if you don't have Mathematica, Maple, or other computer algebra system.

But if one has an existing example like Empy2's identity above where there are obvious patterns in the coefficients, then that helps in reducing the number of unknowns and/or equations.

Answer 2

The answer by @TitoPiezasIII has a long list of comments, so I start another answer.

Here are some similar quadratic sets of answers, but not based on $x,y,x+y$. They are based on other simple solutions.

$$[14(n^2+6)]^4+[3(n^2-6)+34n]^4+[7(n^2-6)+14n]^4=2z^2\\ [3(n^2-20)]^4+[14(n^2+20)+119n]^4+[7(n^2+20)+49n]^4=2z^2\\ [1(n^2-30)]^4+[3(n^2+30)+31n]^4+[10(n^2+30)+109n]^4=2z^2$$

Edit : The form of these solutions, based on a solution to $x^4+y^4+z^4-2w^2$, is $$[x(n^2-b)]^4+[y(n^2+b)+cn]^4+[z(n^2+b)+dn]^4 = \\ 2[w(n^2+b)^2+p(n^2+b)n+qn^2]^2$$ This is a degree-8 polynomial in $n$. Equate coefficients. The nine equations, which connect $b,c,d,p,q$ can be reduced to a single, degree-six polynomial in the variable $c/d$. If this polynomial has a rational root, we get a solution to the quadratic.
For example, if $x=1,y=3,z=10,w=71$, the degree six polynomial has $18$-digit coefficients. It seems to have two rational roots, $31/109$ and $1199/3799$, and four complex ones. The rational roots can be used to produce the third solution above, and $$[1(n^2-35700)]^4+[3(n^2+35700)+1199n]^4+[10(n^2+35700)+3799n]^4=\\ 2[71(n^2+35700)^2+53963(n^2+35700)n+ 10256249n^2]^2$$

Answer 3

$a=55k, b=2k, c=k, d=2139k^2$ ($k\in \mathbb{N}$) can be one example of parametrization such that $a\neq b+c$ or anything else.