The equation $A^4+B^4+C^4=D^4$

Elkies found a solution

$2682440^{4}+15365639^{4}+18796760^{4}=20615673^{4}$

giving a counter examples to Euler’s conjecture that there are no solutions in positive integers to $x_1^n +x_2^n+\cdots x_k^n =y^n, n>2, k\ge 2$ , the generalizations of Fermat’s equations $x^n+y^n =z^n$ to multiple variables .

In the case $n=5$ , a counterexample

$27^{5}+84^{5}+110^{5}+133^{5}=144^{5}$

was found by computer search, but $n=4$ was much harder, and only after Elkies found the above solution, Frye found the following solution by a computer search.

$95800^{4}+217519^{4}+414560^{4}=422481^{4}$

Finding integer solutions to this equation is equivalent to finding rational solutions to

$x^4+y^4+z^4=1$

by the map

$(r,s,t) =(\frac{A}{D}, \frac{B}{D}, \frac{C}{D})$

We now concentrate on finding rational solutions to this equation. The main idea now is to relate to a slightly different surface, use it parametrizations to get parametrization for the surface $x^4+y^4+z^4=1$ . The parametrization reduce the problem to finding solutions on some elliptic curves.

The Surface $r^{4}+s^{4}+t^{4}=1$ can seen as the intersection of the surface $r^{4}+s^{4}+t^{2}=1$ with $t=z^2$ . So we need to find solutions to $r^{4}+s^{4}+t^{2}=1$ where $t$ is a square.

The Surface $r^{4}+s^{4}+t^{2}=1$ :

The surface can be parametrized a pencil of conics (that is for each fixed $u$ ), the solution is a conic -degree two in two variables)

$r=x+y, \quad s=x-y$
$\left(u^{2}+2\right) y^{2}=-\left(3 u^{2}-8 u+6\right) x^{2}-2\left(u^{2}-2\right) x-2 u$
$\left(u^{2}+2\right) t=4\left(u^{2}-2\right) x^{2}+8 u x+\left(2-u^{2}\right)$

We have $u=\frac{-1+(r+s)^{2} \pm t}{r^{2}+r s+s^{2}+r+s}$ .

Also $u \mapsto 2 / u$ changes $(r, s, t, x, y)$ to $(-s,-r,-t,-x, y)$ .

There are many ways to arrive at this parametrization : Bremner wrote $r^4+s^4+t^2=1$ as

$2\left(1+r^{2}\right)\left(1+s^{2}\right)=\left(1+r^{2}+s^{2}\right)^{2}+t^{2}$

and factored the expressions by arithmetic in Gaussian integers. (Zagier, Elkies also found them using different methods)

A rational solution on the surface corresponds to a rational $u$ , and from the above we can assume

$u =\frac{2m}{n}$

for relatively prime odd integers $m \ge 0$ and $n.$

So we have

$r=x+y, \quad s=x-y$
$\left(2 m^{2}+n^{2}\right) y^{2}=-\left(6 m^{2}-8 m n+3 n^{2}\right) x^{2}-2\left(2 m^{2}-n^{2}\right) x-2 m n$
$\left(2 m^{2}+n^{2}\right) t=4\left(2 m^{2}-n^{2}\right) x^{2}+8 m n x+\left(n^{2}-2 m^{2}\right)$

Example:

$(m,n)= (2,1)$ and hence $u=4$ give

$9 y^{2}=-11 x^{2}-14 x-4$

The solution $(x,y) =\left(-\frac{1}{2}, \frac{1}{6}\right)$ gives $(r,s,t) =\left(\frac{1}{3}, \frac{2}{3}, \frac{8}{9}\right)$ .

If you use this one point and draw lines on the plane to find the other intersection points on the conic, we can find the parametrization for $9 y^{2}=-11 x^{2}-14 x-4$ given by

$(x, y)=\left(-\frac{k^{2}+2 k+17}{2 k^{2}+22},-\frac{k^{2}+6 k-11}{6 k^{2}+66}\right)$

which gives

$(r, s, t)=\left(\frac{2 k^{2}+6 k+20}{3 k^{2}+33}, \frac{k^{2}+31}{3 k^{2}+33}, \frac{4\left(2 k^{4}-3 k^{3}+28 k^{2}-75 k+80\right)}{\left(3 k^{2}+33\right)^{2}}\right)$

The Surface $r^{4}+s^{4}+t^{4}=1$ :

Now imposing that $t$ is a square in the previous parametrization, we get the following parametrization for the surface $r^{4}+s^{4}+t^{4}=1$

$r=x+y, \quad s=x-y$
$\left(2 m^{2}+n^{2}\right) y^{2}=-\left(6 m^{2}-8 m n+3 n^{2}\right) x^{2}-2\left(2 m^{2}-n^{2}\right) x-2 m n$ ,
$\pm\left(2 m^{2}+n^{2}\right) t^{2}=4\left(2 m^{2}-n^{2}\right) x^{2}+8 m n x+\left(n^{2}-2 m^{2}\right)$

Example:

$(m,n) =(0,1)$
$y^{2}=-3 x^{2}+2 x$
$\pm t^{2}=1-4 x^{2}$

$(x,y)=(0, 0)$ helps to get the parametrization

$x=\frac{2}{k^{2}+3}, \quad y=k x$

and hence

$\pm t^{2}=1-4 x^{2}=\frac{k^{4}+6 k^{2}-7}{\left(k^{2}+3\right)^{2}}$

With $z=(k^{2}+3) t$

we get

$\pm z^{2}=k^{4}+6 k^{2}-7$

Coordinate changes $k=1-4 /(1 \mp X), z=8 Y /(1 \mp X)^{2}$ gives two Weierstrass equations

$Y^{2}=X^{3}+X \mp 2$

These elliptic curves have only finite many solutions and correspond to the trivial solutions

$(r,s,t) =(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm)$

It can be checked (using quadratic reciprocity and properties of ternary quadratic forms) that the second and third equations for parametrization will have infinitely many solutions $(x,y,t)$ if and only if the squarefree parts of

$2m^2+n^2$ , $2m^2-4mn+n^2$ and $2m^2 \pm 2mn+2n^2$

have all prime factors $1 \mod 8$ .

The smallest $(m,n)$ which satisfy these constraints are

$(0,1), (4,-7),(8,-5),(8,-15),(12,5),(20,-1)$ , and $(20,-9)$ .

$(0,1)$ gave us trivial solutions. $(m,n) =(4,-7)$ doesn’t give solutions (simple local obstructions)

$(m, n) =(8, -5)$ gives

$153 y^{2}=-779 x^{2}-206 x+80$
$\pm 153 t^{2}=412 x^{2}-320 x-103$

The solution $(x, y)=(3 / 14,1 / 42)$ allows us to get the parametrization

$x=\frac{51 k^{2}-34 k-5221}{14\left(17 k^{2}+779\right)}, \quad y=\frac{17 k^{2}+7558 k-779}{42\left(17 k^{2}+779\right)}$

$\pm 21^{2}\left(17 k^{2}+779\right)^{2} t^{2}~=-4\left(31790 k^{4}-4267 k^{3}+1963180 k^{2}-974003 k-63237532\right)$

Looking at $\bmod ~3$ , we see that only the plus sign is valid.

Now the change of variables,

$X=(k+2) / 7, Y=3\left(17 k^{2}+779\right) t / 14$

gives the elliptic curve

$Y^{2}=-31790 X^{4}+36941 X^{3}-56158 X^{2}+28849 X+22030$

Computer search gave the following solution:

$(X, Y)=\left(-\frac{31}{467}, \frac{30731278}{467^{2}}\right)$

which corresponds to

$(r, s, t)=\left(-\frac{18796760}{20615673}, \frac{2682440}{20615673}, \frac{15365639}{20615673}\right)$

that is

$2682440^{4}+15365639^{4}+18796760^{4}=20615673^{4} .$

More solutions: Using the solutions $P_{+}=(X, Y)=\left(-\frac{31}{467}, \frac{30731278}{467^{2}}\right)$ and $P_{-}=(X, -Y)=\left(-\frac{31}{467}, -\frac{30731278}{467^{2}}\right)$ on the elliptic curve $Y^{2}=-31790 X^{4}+36941 X^{3}-56158 X^{2}+28849 X+22030$ we can generate more solutions by group law on the curve. For instance $P_{+}-P_{-}$ gives the following solution

$A =1439965710648954492268506771833175267850201426615300442218292336336633$
$B =4417264698994538496943597489754952845854672497179047898864124209346920$
$C =9033964577482532388059482429398457291004947925005743028147465732645880$
$D=9161781830035436847832452398267266038227002962257243662070370888722169$