Near to Euler’s 4th power taxicab equation solution using $W^{4}+X^{4}=Y^{2}+Z^{4}$?
问题内容
The above given equation solution is very easy just make it to an elliptic curve. For $$ W^{4}+X^{4}=Y^{2}+Z^{4} $$ Divide both sides $Z^{4}$, we wil get $$ \left(\frac {W}{Z}\right)^4 + \left(\frac{X}{Z}\right)^4 = \left(\frac{Y}{Z^2}\right)^2 +1 $$ If we substitute $\frac{W}{Z} = (u+v)$, $\frac{X}{Z} = (u-v)$, $\frac{Y}{Z^2}= w$ then we will get $$ w^{2}=2u^4+12u^2v^2+2v^4-1 $$
Now put $v=2$, then $w^2= 2u^4+48u^2+31$,
(u,w) = (1,9) (-1,9) (1,-9) (-1,-9)
we can easily make this into elliptic curve Weierstrass form and use group law to generate infinitely many solutions.
My question is that if we make $Y= a$ square then we can get Euler’s famous 4th power taxicab number solution. I am looking for such a method e.g. let $Y= t^2$ is expressed by another quadratic , cubic or biquadratic polynomial that can make a square. I had heard about k3 surface but I have not very clear conception of this topic. Is this problem related to k3 surface? I also heard that pell equation also involved to find Euler 4th power taxicab number solution. This is a different method. Any good suggestion will be appreciated.
回答 (1)
There are four ways you can use this identity (I think it is by Gerardin):
$$(17p^2-12pq-13q^2)^4 + (17p^2+12pq-13q^2)^4 = (289p^4+14p^2q^2-239q^4)^2 + (17p^2-q^2)^4$$
I. 1st way: By solving an elliptic curve, you can get infinitely many solutions to $289p^4+14p^2q^2-239q^4 = r^2$, hence
$$x^4+y^4=z^4+t^4$$
Example. Let $(p,q)=(33,17)$ or $(p,q)=(11,3)$ or $(p,q)=(16,17)$ and, after removing common factors, you get the three smallest solutions,
$$59^4 + 158^4 = 133^4 + 134^4$$ $$193^4 + 292^4 = 257^4 + 256^4$$ $$157^4 + 227^4 = 7^4 + 239^4$$
and infinitely more.
II. 2nd way: By solving the Pell equation $17p^2-q^2=1$, then you get infinitely many solutions to,
$$x^4+y^4=z^2+1$$
Note: Fermat proved there are no solutions to $x^4+y^4=z^2$ so this is a "near-miss".
III. 3rd way: By solving $17p^2\pm12pq-13q^2 = \square$, then you get infinitely many solutions to,
$$x^\color{red}8+y^\color{red}8=z^2+t^4$$
Example: Let $(p,q) = (7,4)$ to get,
$$17^\color{red}{8} + 31^\color{red}8 = 643681^2 + 817^4$$
IV. 4th way: By solving an elliptic curve, you can get infinitely many solutions to $17p^2-12pq-13q^2 = r^4$, hence
$$x^\color{red}{16}+y^4=z^2+t^4$$
Example: Let $(p,q) = (31,4)$ to get,
$$11^\color{red}{16} + 17617^4 = 267051649^2 + 16321^4$$
P.S. This identity relies on a special solution to the Pell equation $x^2-2y^2=-1$, namely $(x,y)=(239,13^2)$, so it may be hard to find similar ones.