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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.

a-lawliet
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2 Answers2

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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:$(x,y,z)=(-11/93,30/93,92/93)$

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Sam
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    very interesting reference (+1). However how one can show from that that OP's system is missing some solutions ? – G Cab Jun 04 '18 at 23:26
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    Sam copied wrong a couple of signs, that's why the particular solution is wrong, but the actual formulas on the paper work. However, how do you know my formulas are missing solutions? And, in that case, how could I see that? I am not asking for an exhaustive guide, just some idea I can work with. – a-lawliet Jun 07 '18 at 11:21
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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.

Sam
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