Intereting Posts

Mapping $\Delta(2,2,2)\mapsto \Delta(4,4,2)$…
Is there a Dihedral group of order 4?
If $m=\operatorname{lcm}(a,b)$ then $\gcd(\frac{m}{a},\frac{m}{b})=1$
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$
Are these zeros equal to the imaginary parts of the Riemann zeta zeros?
Number of solutions to $a_1 + a_2 + \dots + a_k = n$ where $n \gt 0$ and $0 \lt a_1 \leq a_2 \leq \dots \leq a_k$ are integers.
Uniformly distributed rationals
If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected?
3 random numbers to describe point on a sphere
If $f$ is ananlytic in $D_r(z_0)$\ {$z_0$} and $Ref(z)>0$ for all $z\in D_r(z_0)$\ {$z_0$} then $z_0$ is a removable singularity
Calculate an integral in a measurable space
Find the trace of the matrix?
When is a flat morphism open?
Exactly who popularized the modern definition of domain and codomain of functions?
Alternative proof that Harmonic sum is not an integer

While answering this question, I got myself stumped with this crazy system with an evil graph: $$\begin{cases}

18xy^2+x^3=12 \\

27x^2y+54y^3=38

\end{cases}$$ 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.

- Solve the System of Equations in Real $x$,$y$ and $z$
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- How to solve for any given natural number n?
- How to find the complex roots of $y^3-\frac{1}{3}y+\frac{25}{27}$
- What is the solution to the system $\frac{df_n}{dt} = kf_{n-1}-(k+l)f_n+lf_{n+1}$?
- Proving that two systems of linear equations are equivalent if they have the same solutions
- Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double real root.
- Solving Cubic when There are Known to be 3 Real Roots
- Blowing-up a singular point
- Is there a simpler approach to these system of equations?

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{cases}

uv^2+2u^3=3\\

18u^2v+v^3=19\\

\end{cases}$$

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

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

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

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

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

**Added later**: Just to elaborate, the given equations change into something of the form

$$\begin{cases}

ru^3+(1-r)uv^2=1\\

sv^3+(1-s)u^2v=1\\

\end{cases}$$

This leads to

$$(u-v)\left(ru^2-(1-r-s)uv+sv^2\right)=0$$

and the quadratic contributes nothing to the (real) solution if $(1-r-s)^2\lt4rs$.

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 $y=19x/(6(2x^3+3)$. This in turn gives the following result:

$$

(x-2)(x^2 + 2x + 4)(2x^2 + 4x + 3)(4x^4 – 8x^3 + 10x^2 – 12x + 9)=0

$$

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

I see that you used *Mathematica* to find your “evil graph”. For a result with no clutter, you can also use the command,

```
Factor[Resultant[18y^2x+x^3-12, 27x^2y+54y^3-38,y]]
```

which will eliminate the variable *y* and give you Burde’s neat result.

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