Generalized 4-variable Diophantine equation

Question

The diophantine equation $(a^2+b^2+c^2+d^2)^2 = 2(a^4+b^4+c^4+d^4)$, which it's the case of Descartes' circle theorem where all curvature are integer perfect squares, was cleverly solved by Euler noticing some pythagorean triples. Generalising, if instead of 2 there was another number, is it still possible to use some similar trick? In particular, it's possible to find a parametrization to the primitive integer solutions of $(a^2+b^2+c^2+d^2)^2 = 3(a^4+b^4+c^4+d^4)$? Thanks in advance.

Answer 1

Euler found that the equation,

$$(v^2 + x^2 + y^2 + z^2)^2 = 2(v^4 + x^4 + y^4 + z^4)\tag1$$

is equivalent to any three of the Pythagorean triples,

$$(2v x)^2 + (2y z)^2 = (v^2 + x^2 - y^2 - z^2)^2\\ (2v y)^2 + (2x z)^2 = (v^2 - x^2 + y^2 - z^2)^2\\ (2v z)^2 + (2x y)^2 = (v^2 - x^2 - y^2 + z^2)^2$$

The reference for this is Dickson's "History of the Theory of Numbers" though I've forgotten the specific page. This equivalence used to be in the Wikipedia article on Descartes' theorem, but someone deleted it. There are parametric solutions to $(1)$.

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Solution 1

If $a+b = c$ and $d = 2t$, then the equation simplifies as,

$$a^2+ab+b^2 = t^2$$

which is the same condition as in Tomita's answer.

Solution 2

If we prefer $a+b \neq c$, then there is also a parametrization using, perhaps not surprisingly, Pythagorean triples. Let,

$$(v,\,x,\,y,\,z) = \big(a e - b d,\, (a - d) (a + d),\, a(c + f),\, d(c + f)\big)$$

But we must first solve the pair,

$$a^2+b^2 =c^2\\ \,d^2+e^2 = f^2\\ \qquad ab = de$$

a simple solution of which is,

$$\big(a,b,c,d,e,f\big) = \big(u^2-v^2,\, 2uv,\, u^2+v^2,\, u^2-w^2,\, 2uw,\, u^2+w^2\big)$$

where $(u,v,w)$ obey the equation,

$$v^2+vw+w^2 = u^2$$

the same condition as Solution 1 (with different variables).

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P.S. A quick search of,

$$\frac{(a^2 + b^2 + c^2 + d^2)^2} {a^4 + b^4 + c^4 + d^4} = N$$

with $a<b<c<d<100$ yields no other integer $N$ except $N=2,3$ which seems to confirm David's comment (at least within that limited range).

Answer 2

$$(a^2+b^2+c^2+d^2)^2=3(a^4+b^4+c^4+d^4)$$ Let assume $c=a+b,$ then we get $$a^2+ab+b^2=d^2$$ A parametric solution is given $$(a,b,c,d)=(p^2-q^2, 2pq+q^2, p^2+2pq, q^2+pq+p^2)$$ where $p,q$ are any integer.