On a remarkable system of fourth powers using $x^4+y^4+(x+y)^4=2z^4$ The problem is to find four integers $a,b,c,d$ such that,
$$a^4+b^4+(a+b)^4=2{x_1}^4\\a^4+c^4+(a+c)^4=2{x_2}^4\\a^4+d^4+(a+d)^4=2{\color{blue}{x_3}}^4\\b^4+c^4+(b+c)^4=2{x_4}^4\\b^4+d^4+(b+d)^4=2{x_5}^4\\c^4+d^4+(c+d)^4=2{x_6}^4$$
As W. Jagy pointed out, the form $x^4+y^4+(x+y)^4 = 2z^4$ appear in the context of triangles with integer sides and one $120^\circ$ angle. PM 2Ring discovered that, remarkably, the quadruple,
$$a,b,c,d = 195, 264, 325, 440$$
yields five integer $x_i$ (except $x_3$).
I found that, using an elliptic curve, it can be showed there are infinitely many non-zero integer triples with $\gcd(a,b,c)=1$ such that three $x_i$ are integers. 

Q: However, are there infinitely many quadruples with $\gcd(a,b,c,d)=1$ such that at least five of the $x_i$ are integers?

 A: After mulling over the problem, it turned out the same elliptic curve can make five of the $x_i$ as integers. To start, note that,
$$x^4+y^4+(x+y)^4 = 2(x^2+xy+y^2)^2$$
Thus, the system is reduced to finding,
$$\color{blue}{a^2+ab+b^2 = x_1^2}\tag1$$
$$\color{blue}{a^2+ac+c^2 = x_2^2}\tag2$$
$$b^2+bc+c^2 = x_4^2\tag3$$
$$\color{brown}{b^2+bd+d^2 = x_5^2}\tag4$$
$$\color{brown}{c^2+cd+d^2 = x_6^2}\tag5$$
Assume $b,c$ as constant. Plugging them into $(1),(2)$, then a pair $\color{blue}{P_1}$ of quadratics in $\color{blue}a$ must be made a square. Since this has a rational point, it is birationally equivalent to an elliptic curve. 
Plugging $b,c$ into $(4),(5)$, a pair $\color{brown}{P_2}$ of quadratics in $\color{brown}d$ must be made a square. But $P_1$ and $P_2$ have the same form. Thus, two different points in the same elliptic curve will yield the $a,d$. Clearing denominators, we then get another quadruple,
$$a,b,c,d = 232538560625,\, 670011598080,\, 824824884000,\, 749417043168$$
and an infinite more with $\gcd(a,b,c,d)=1$. (Presumably, smaller ones may exist.)
