On the complete rational solutions of $x^3+y^3+z^3=1$

Question

I have to check that the rational solutions of $x^3+y^3+z^3=1$ are given by giving rationals values to $(s,t)$ at the formulas:

$$x(s,t)=\frac{3t-\frac{1}{3}(s^2+st+t^2)^2}{t(s^2+st+t^2)-3}$$

$$y(s,t)=\frac{3s+3t+\frac{1}{3}(s^2+st+t^2)^2}{t(s^2+st+t^2)-3}$$

$$z(s,t)=\frac{-3-(s^2+st+t^2)(s+t)}{t(s^2+st+t^2)-3}$$

After some dirty work, I can see that these expressions for $x,y,z$ satisfy the equation $x^3+y^3+z^3=1$, but I still have to see that all the rational solutions are of that form, and I don't know how to approach this (it is necessary, because there could be other solutions that can't be expressed as above, and then these formulas wouldn't be a complete solution).

Previously to this, I have found explicit expressions for the rational solutions of $x^2+y^2+z^2=1$ and the more general equation $x_1^2+...+x_n^2=1$, both through the stereographic projection. I have tried to do something similar with the cubic equation but it doesn't seem to work. I tried to transform it to the form $x^3+y^3+z^3=t^3$ and find the integer solutions to this, but I couldn't and I hardly found information about it on internet, and the little I've found doesn't match with the family of solutions I am working with (which is the one I'm interested in)

I genuinely don't know how to approach this, so I would appreciate any idea. Thanks.

Answer 1

Above equation shown below:

$x^3+y^3+z^3=1$ -----$(1)$

Solution shown by (OP) is not a general solution. Hence he can refer to paper by Mr. Ajai Choudhry who arrived at a general solution in his paper shown at the link below,

https://projecteuclid.org/euclid.rmjm/1181071714#export

Above eqn. (1) has solution in the said paper & is shown below:

$dp = c(-a^3- b^3+ c^3 )$

$dq = (-a^4+ 2a^3 b- 3a^2 b^2+ 2ab^3- b^4 )+ (a+ b) c^3$

$dr = (a^4- 2a^3 b + 3a^2 b^2- 2ab^3+ b^4 )+ (2a- b) c^3$

---

$dt = c(a^3+(a-b)^3+ c^3 )$

Condition : $ c^3 > (a^3+ b^3) $ and 'd' is a constant of multiplication

And $(x,y,z)=(p/t,q/t,r/t)$

For $(a,b,c,d)=(2,1,3,1)$ we get:$(x,y,z)=(-11/93,30/93,92/93)$

Answer 2

Re: Adam Bailey & Ai Yagami. I have corrected the typos that you'll have mentioned.

Regarding query by Ai Yagami about eqn.

$x^3+y^3+z^3=1$ -----(1)

Since $(x,y,z)=(3/6,4/6,5/6)$ is a solution to equation (1), than substituting the numerical solution in equation given by (OP) & solving for (s & t) we get the below mentioned equation:

$3=((s^2+st+t^2)(s+3t))/(s+2t+1)$ ----(2)

Above eqn. (2) has solution $(s, t)=(1,1)$

But, equation given by (OP) is not satisfied at (s,t)=(1,1). Hence it is a contradiction.