Parametrizing the solutions to a diophantine equation of degree four

Question

Good evening,

Consider $x^4+y^4+z^4=2t^4$ where x,y,z,t integer.

Is it known how to find all parametrisation of this equation ?

If you have any parametrisation or reference of this equation, please post it

Thank you.

Answer 1

Ramanujan gave two parametrizations: if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(listed in Mathworld, equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract] [summary]

The F. Ferrari Identity (1909) gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also K. Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

For $k=2$ and $k=4$, the Ford parametrizations are the same as the two Ramanujan ones. For $k=2$, this is simple, and for $k=4$, $S_0 = a^4-4 a b^3$, $S_1 = b^4-4 a^3 b$, $S_2 = 6 a^2 b^2$, and

$$\begin{align} S_0 - S_1 = -(c^3+2abc)(a-b)\\ S_1 - S_2 = -(b^3+2abc)(c-a)\\ S_2 - S_0 = -(a^3+2abc)(b-c)\\ \end{align}$$

where $c=-a-b$.

Edit: A complete discussion of these identities, by S. Ramanujan, F. Ferrari and Kevin Ford, plus further identities by S. Bhargava, may be found on pages 96 to 101 of Bruce Berndt's Ramanujan's Notebooks, Part IV (1994), at e.g. http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20IV.pdf.

Answer 2

What exactly do you mean by "paramatrization"? Anyway, since your equation is homogeneous, you're really asking for $\mathbb Q$-rational points on the surface $S$ in $\mathbb P^3$ defined by your equation. The surface $S$ is an example of what is known as a K3 surface. Most likely what you have in mind for a paramatrization would be homogeneous polynomials $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb Z[u,v,w]$ so that you get all solutions by plugging in integer values of $u$, $v$, and $w$. Such a solution very likely does not exist. Here's why. First, there is no way to parametrize the complex solutions using polynomial functions $x(u,v,w),y(u,v,w),z(u,v,w),t(u,v,w)\in\mathbb C[u,v,w]$, because a K3 surface is not birationally equivalent to $\mathbb P^2$. So that means there are two possibilities. First, the integer solutions are Zariski dense, in which case you won't have a parametrization. Second, the integer solutions lie on a finite number of curves. (I don't know offhand how to distinguish between these two possibilities.) However, is is conjectured that for any K3 surface, there is a finite extension $K/\mathbb Q$ such that the $K$-rational points are Zariski dense. So if it happens that you can parametrize the integer solutions to your particular equation, that's in some sense because your field $\mathbb Q$ isn't big enough to really reflect the arithmetic structure of the problem. And the solutions won't be that interesting, since they'd only be a finite number of one-parameter families.