Noam Elkies found that $x^4+y^4+z^4 = 1$ has infinitely many rational points $xyz \neq 0$ using an elliptic curve. We use a different approach that will produce pairs of solutions and a parametric elliptic curve. Let,

$$x^4+y^4+z^4 = 1\tag1$$ $$\frac{(x-y)^2-z^2-1}{x^2-xy+y^2+(x-y)}=u\tag2$$ $$\frac{(y-z)^2-x^2-1}{y^2-yz+z^2+(y-z)}=v\tag3$$ $$\frac{(z-x)^2-y^2-1}{z^2-zx+x^2+(z-x)}=w\tag4$$

where the $(u,v,w)$ have the nice relationship,

$$2(u+v+w)-uvw-4=0$$

As such, $w$ is linearly dependent on $(u,v)$ and $(4)$ is redundant. So we use the first three equations to solve for the three unknowns $(x,y,z)$. It turns out each are roots of quadratics and we get a solution of form,

$$\big(a_1\pm a_2\sqrt{D^2}\big)^4+\big(a_3\pm a_4\sqrt{D^2}\big)^4+\big(a_5\pm a_6\sqrt{D^2}\big)^4 = 1$$

For those interested in the complicated expressions for the $a_i$, they can be found here. But the important variable is the quartic to be made a square,

$$D^2 = 4(-6 - 2u + u^2)(2 - 2u + u^2) - 8(-2 - 4u + u^2)(2 - 2u + u^2)v - 16u(4 - 3u + u^2)v^2 - 4(4 - 12u + 4u^2 - 2u^3 + u^4)v^3 + (4 - 8u - 4u^3 + u^4)v^4$$

If there is rational $(u,v,D)$, then it is birationally equivalent to an elliptic curve. And if we consider only $u$ of small height (numerator and denominator with absolute value less than $1000$), then only $17 $ $u$ are known. These are,

$$u = -\dfrac{9}{20},\, -\dfrac{29}{12},\, -\dfrac{93}{80},\, -\dfrac{400}{37},\, -\dfrac{136}{133},\, \dfrac{201}{4},\, -\dfrac{5}{8},\, -\dfrac{477}{692}$$ $$u = -\dfrac{41}{36},\,-\dfrac{5}{44},\, \dfrac{233}{60},\, -\dfrac{56}{165},\, -\dfrac{125}{92},\, -\dfrac{361}{540},\, -\dfrac{817}{660},\, -\dfrac{865}{592},\, \dfrac{553}{80}$$

Question: Are there ways to find more $u$ of similar height other than brute-force? (The last was found by D. Fulea.)

P.S. In the link above, a brute-force search was done by S. Tomita and D. Fulea who also found,

$$u = \dfrac{1744}{495},\, -\dfrac{1376}{705},\, -\dfrac{3168}{1553},\, \dfrac{1873}{200}$$

Surely the $17$ can't be the only ones of small height? To illustrate, Bremner found four $u$ back in 2015 including $(u,v) = -\frac{5}{44},\, \frac{57878913}{12642040}$ and I'm not sure if he used brute force or an efficient algorithm. So maybe there are more "small" $u$ but the bounds used for $v$ was not high enough? How do we efficiently find such $u$? (The focus on small $u$ is they tend to produce $a^4+b^4+c^4 = d^4$ where $d$ is "smallish".)

Noam Elkies found that $x^4+y^4+z^4 = 1$ has infinitely many rational points $xyz \neq 0$ using an elliptic curve. We use a different approach that will produce pairs of solutions and a parametric elliptic curve. Let,

$$x^4+y^4+z^4 = 1\tag1$$ $$\frac{(x-y)^2-z^2-1}{x^2-xy+y^2+(x-y)}=u\tag2$$ $$\frac{(y-z)^2-x^2-1}{y^2-yz+z^2+(y-z)}=v\tag3$$ $$\frac{(z-x)^2-y^2-1}{z^2-zx+x^2+(z-x)}=w\tag4$$

where the $(u,v,w)$ have the nice relationship,

$$2(u+v+w)-uvw-4=0$$

As such, $w$ is linearly dependent on $(u,v)$ and $(4)$ is redundant. So we use the first three equations to solve for the three unknowns $(x,y,z)$. It turns out each are roots of quadratics and we get a solution of form,

$$\big(a_1\pm a_2\sqrt{D^2}\big)^4+\big(a_3\pm a_4\sqrt{D^2}\big)^4+\big(a_5\pm a_6\sqrt{D^2}\big)^4 = 1$$

Those interested in the complicated expressions for the $a_i$, they can be found here. But the important variable is the quartic to be made a square,

$$D^2 = 4(-6 - 2u + u^2)(2 - 2u + u^2) - 8(-2 - 4u + u^2)(2 - 2u + u^2)v - 16u(4 - 3u + u^2)v^2 - 4(4 - 12u + 4u^2 - 2u^3 + u^4)v^3 + (4 - 8u - 4u^3 + u^4)v^4$$

If there is rational $(u,v,D)$, then it is birationally equivalent to an elliptic curve. And if we consider only $u$ of small height (numerator and denominator with absolute value less than $1000$), then only $17 $ $u$ are known. These are,

$$u = -\dfrac{9}{20},\, -\dfrac{29}{12},\, -\dfrac{93}{80},\, -\dfrac{400}{37},\, -\dfrac{136}{133},\, \dfrac{201}{4},\, -\dfrac{5}{8},\, -\dfrac{477}{692}$$ $$u = -\dfrac{41}{36},\,-\dfrac{5}{44},\, \dfrac{233}{60},\, -\dfrac{56}{165},\, -\dfrac{125}{92},\, -\dfrac{361}{540},\, -\dfrac{817}{660},\, -\dfrac{865}{592},\, \dfrac{553}{80}$$

Question: Are there ways to find more $u$ of similar height other than brute-force? (The last was found by D. Fulea.)

P.S. In the link above, a brute-force search was done by S. Tomita and D. Fulea who also found,

$$u = \dfrac{1744}{495},\, -\dfrac{1376}{705},\, -\dfrac{3168}{1553},\, \dfrac{1873}{200}$$

Surely the $17$ can't be the only ones of small height? To illustrate, Bremner found four $u$ back in 2015 including $(u,v) = -\frac{5}{44},\, \frac{57878913}{12642040}$ and I'm not sure if he used brute force or an efficient algorithm. So maybe there are more "small" $u$ but the bounds used for $v$ was not high enough. So how do we efficiently find such $u$?

The focus on small $u$ is they tend to produce $a^4+b^4+c^4 = d^4$ where $d$ is "smallish". For example, using Fulea's recent discovery of $(u,v) = \frac{553}{80},-\frac{33400}{19537}$, substituting it into the system, yields a pair of $(x,y,z)$ the smaller one is,

$$24743080^4 + 3971389576^4 + 4657804375^4 = 5179020201^4$$

which at just $d\approx 5.1\times10^9$ is the smallest known since 2008.

Noam Elkies found that $x^4+y^4+z^4 = 1$ has infinitely many rational points $xyz \neq 0$ using an elliptic curve. We use a different approach that will produce pairs of solutions and a parametric elliptic curve. Let,

$$x^4+y^4+z^4 = 1\tag1$$ $$\frac{(x-y)^2-z^2-1}{x^2-xy+y^2+(x-y)}=u\tag2$$ $$\frac{(y-z)^2-x^2-1}{y^2-yz+z^2+(y-z)}=v\tag3$$ $$\frac{(z-x)^2-y^2-1}{z^2-zx+x^2+(z-x)}=w\tag4$$

where the $(u,v,w)$ have the nice relationship,

$$2(u+v+w)-uvw-4=0$$

As such, $w$ is linearly dependent on $(u,v)$ and $(4)$ is redundant. So we use the first three equations to solve for the three unknowns $(x,y,z)$. It turns out each are roots of quadratics and we get a solution of form,

$$\big(a_1\pm a_2\sqrt{D^2}\big)^4+\big(a_3\pm a_4\sqrt{D^2}\big)^4+\big(a_5\pm a_6\sqrt{D^2}\big)^4 = 1$$

For those interested in the complicated expressions for the $a_i$, they can be found here. But the important variable is the quartic to be made a square,

$$D^2 = 4(-6 - 2u + u^2)(2 - 2u + u^2) - 8(-2 - 4u + u^2)(2 - 2u + u^2)v - 16u(4 - 3u + u^2)v^2 - 4(4 - 12u + 4u^2 - 2u^3 + u^4)v^3 + (4 - 8u - 4u^3 + u^4)v^4$$

If there is rational $(u,v,D)$, then it is birationally equivalent to an elliptic curve. And if we consider only $u$ of small height (numerator and denominator with absolute value less than $1000$), then only $17 $ $u$ are known. These are,

$$u = -\dfrac{9}{20},\, -\dfrac{29}{12},\, -\dfrac{93}{80},\, -\dfrac{400}{37},\, -\dfrac{136}{133},\, \dfrac{201}{4},\, -\dfrac{5}{8},\, -\dfrac{477}{692}$$ $$u = -\dfrac{41}{36},\,-\dfrac{5}{44},\, \dfrac{233}{60},\, -\dfrac{56}{165},\, -\dfrac{125}{92},\, -\dfrac{361}{540},\, -\dfrac{817}{660},\, -\dfrac{865}{592},\, \dfrac{553}{80}$$

Question: So how do find more $u$ of similar height? (The last was found by D. Fulea.)

P.S. In the link above, a brute-force search was done by S. Tomita and D. Fulea who also found,

$$u = \dfrac{1744}{495},\, -\dfrac{1376}{705},\, -\dfrac{3168}{1553},\, \dfrac{1873}{200}$$

Surely the $17$ can't be the only ones of small height? To illustrate $(u,v) = -\frac{5}{44},\, \frac{57878913}{12642040}$ was found by Andrew Bremner back in 2015. So maybe there are more "small" $u$ but the bounds used for $v$ was not high enough? How do we efficiently find such $u$? (Small $u$ tend to produce $a^4+b^4+c^4 = d^4$ where $d$ is "smallish".)

Noam Elkies found that $x^4+y^4+z^4 = 1$ has infinitely many rational points $xyz \neq 0$ using an elliptic curve. We use a different approach that will produce pairs of solutions and a parametric elliptic curve. Let,

$$x^4+y^4+z^4 = 1\tag1$$ $$\frac{(x-y)^2-z^2-1}{x^2-xy+y^2+(x-y)}=u\tag2$$ $$\frac{(y-z)^2-x^2-1}{y^2-yz+z^2+(y-z)}=v\tag3$$ $$\frac{(z-x)^2-y^2-1}{z^2-zx+x^2+(z-x)}=w\tag4$$

where the $(u,v,w)$ have the nice relationship,

$$2(u+v+w)-uvw-4=0$$

As such, $w$ is linearly dependent on $(u,v)$ and $(4)$ is redundant. So we use the first three equations to solve for the three unknowns $(x,y,z)$. It turns out each are roots of quadratics and we get a solution of form,

$$\big(a_1\pm a_2\sqrt{D^2}\big)^4+\big(a_3\pm a_4\sqrt{D^2}\big)^4+\big(a_5\pm a_6\sqrt{D^2}\big)^4 = 1$$

For those interested in the complicated expressions for the $a_i$, they can be found here. But the important variable is the quartic to be made a square,

$$D^2 = 4(-6 - 2u + u^2)(2 - 2u + u^2) - 8(-2 - 4u + u^2)(2 - 2u + u^2)v - 16u(4 - 3u + u^2)v^2 - 4(4 - 12u + 4u^2 - 2u^3 + u^4)v^3 + (4 - 8u - 4u^3 + u^4)v^4$$

If there is rational $(u,v,D)$, then it is birationally equivalent to an elliptic curve. And if we consider only $u$ of small height (numerator and denominator with absolute value less than $1000$), then only $17 $ $u$ are known. These are,

$$u = -\dfrac{9}{20},\, -\dfrac{29}{12},\, -\dfrac{93}{80},\, -\dfrac{400}{37},\, -\dfrac{136}{133},\, \dfrac{201}{4},\, -\dfrac{5}{8},\, -\dfrac{477}{692}$$ $$u = -\dfrac{41}{36},\,-\dfrac{5}{44},\, \dfrac{233}{60},\, -\dfrac{56}{165},\, -\dfrac{125}{92},\, -\dfrac{361}{540},\, -\dfrac{817}{660},\, -\dfrac{865}{592},\, \dfrac{553}{80}$$

Question: Are there ways to find more $u$ of similar height other than brute-force? (The last was found by D. Fulea.)

P.S. In the link above, a brute-force search was done by S. Tomita and D. Fulea who also found,

$$u = \dfrac{1744}{495},\, -\dfrac{1376}{705},\, -\dfrac{3168}{1553},\, \dfrac{1873}{200}$$

Surely the $17$ can't be the only ones of small height? To illustrate, Bremner found four $u$ back in 2015 including $(u,v) = -\frac{5}{44},\, \frac{57878913}{12642040}$ and I'm not sure if he used brute force or an efficient algorithm. So maybe there are more "small" $u$ but the bounds used for $v$ was not high enough? How do we efficiently find such $u$? (The focus on small $u$ is they tend to produce $a^4+b^4+c^4 = d^4$ where $d$ is "smallish".)