Arraryeldergox2

2022-06-26

Solving the system

$\{\begin{array}{l}18x{y}^{2}+{x}^{3}=12\\ 27{x}^{2}y+54{y}^{3}=38\end{array}$

and I wonder whether there is some slick method to find the only real root $(x,y)=(2,1/3)$ without relying on Cardano's formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I'm tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

$\{\begin{array}{l}18x{y}^{2}+{x}^{3}=12\\ 27{x}^{2}y+54{y}^{3}=38\end{array}$

and I wonder whether there is some slick method to find the only real root $(x,y)=(2,1/3)$ without relying on Cardano's formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I'm tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

Quinn Everett

Beginner2022-06-27Added 23 answers

It makes sense to change variables, letting $v=3y$ and $x=2u$, so that what we're looking for is $(u,v)=(1,1)$. The equations become

$\{\begin{array}{l}u{v}^{2}+2{u}^{3}=3\\ 18{u}^{2}v+{v}^{3}=19\end{array}$

Multiplying the first by $19$ and the second by $3$ and subtracting gives

$38{u}^{3}-54{u}^{2}+19u{v}^{2}-3{v}^{3}=(u-v)(38{u}^{2}-16uv+3{v}^{2})=0$

The quadratic term is easily seen to be non-negative:

$9{v}^{2}-48uv+114{u}^{2}=(3v-8u{)}^{2}+50{u}^{2}$

Hence we must have $u=v$. Plugging this into either equation gives the desired result, $u=v=1$.

Just to elaborate, the given equations change into something of the form

$\{\begin{array}{l}r{u}^{3}+(1-r)u{v}^{2}=1\\ s{v}^{3}+(1-s){u}^{2}v=1\end{array}$

This leads to

$(u-v)(r{u}^{2}-(1-r-s)uv+s{v}^{2})=0$

and the quadratic contributes nothing to the (real) solution if $(1-r-s{)}^{2}<4rs$

$\{\begin{array}{l}u{v}^{2}+2{u}^{3}=3\\ 18{u}^{2}v+{v}^{3}=19\end{array}$

Multiplying the first by $19$ and the second by $3$ and subtracting gives

$38{u}^{3}-54{u}^{2}+19u{v}^{2}-3{v}^{3}=(u-v)(38{u}^{2}-16uv+3{v}^{2})=0$

The quadratic term is easily seen to be non-negative:

$9{v}^{2}-48uv+114{u}^{2}=(3v-8u{)}^{2}+50{u}^{2}$

Hence we must have $u=v$. Plugging this into either equation gives the desired result, $u=v=1$.

Just to elaborate, the given equations change into something of the form

$\{\begin{array}{l}r{u}^{3}+(1-r)u{v}^{2}=1\\ s{v}^{3}+(1-s){u}^{2}v=1\end{array}$

This leads to

$(u-v)(r{u}^{2}-(1-r-s)uv+s{v}^{2})=0$

and the quadratic contributes nothing to the (real) solution if $(1-r-s{)}^{2}<4rs$

anginih86

Beginner2022-06-28Added 8 answers

Indeed, all rational solutions are given by $(x,y)=(2,1/3)$. We can see this without any theory on elliptic curves, but just with a computation: Substitute ${y}^{2}=(12-{x}^{3})/(18x)$ from the first equation into the second to obtain $y=19x/(6(2{x}^{3}+3)$. This in turn gives the following result:

$(x-2)({x}^{2}+2x+4)(2{x}^{2}+4x+3)(4{x}^{4}-8{x}^{3}+10{x}^{2}-12x+9)=0$

Obviously, only $x=2$ is a rational solution for $x$. Also we see, that it is the only real solution.

$(x-2)({x}^{2}+2x+4)(2{x}^{2}+4x+3)(4{x}^{4}-8{x}^{3}+10{x}^{2}-12x+9)=0$

Obviously, only $x=2$ is a rational solution for $x$. Also we see, that it is the only real solution.

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