Solving the system $x^2 + y^2 = 10$, $x^3 + y^3 = 28$

Question

$$\begin{align} x^2 + y^2 &= 10 \\ x^3 + y^3 &= 28 \end{align}$$

It seems simple, but I can't solve it. I tried to decompose the $x^3 + y^3$, but this doesn't help.

Answer 1

How about $x^2 = a$ and $y^2 = b$

$x^2 + y^2 = a + b = 10$

$x^3 + y^3 = ax + by = 28$

Cubic and quadratic reduced to linearity. Ab hic I believe we need granuma (seeds), values for $a$ and $b$ to make the equations solvable. Do notice that that was probably what was intended, only $2$ equations were provided.

We could also do this: $x^2 = 10 - y^2$. Then ...

$x(10 - y^2) + y^3 = 28$

And then ...

$x = \frac{28 - y^3}{10 - y^2}$.

The two equations then morph into ...

$\left( \frac{28 - y^3}{10 - y^2} \right)^2+ y^2 = 10$

and

$\left( \frac{28 - y^3}{10 - y^2} \right)^3 + y^3 = 28$

Answer 2

Obviously you're supposed to test small numbers until you solve it, but assuming we don't want to do that, you can argue as follows. We can cube the first equation and square the second one to get $$x^6 + 3x^2y^2(x^2 + y^2) + y^6 = 1000$$ $$x^6 + 2x^3y^3 + y^6 = 784$$ Substitute $x^2 + y^2 = 10$ into the first equation and then subtract the second equation from the first. You get $$3(x^2y^2) - 2(x^3y^3) = 216$$ Letting $z = xy$ this is the same as $$2z^3 - 3z^2 + 216 = 0$$ Trying rational roots, we'll eventually get $z = 3$ as a solution, and factoring out $z - 3$ we get $$(z - 3)(z^2 - 12z + 36) = 0$$ We can then solve the quadratic, and we get 3 possible values of $z$: $z = 3, z = -6 + 6\sqrt{2},$ and $z = -6 - 6\sqrt{2}$.

Next, we write the second given equation as $$(x + y)(x^2 - xy + y^2) = 28$$ Since we know $x^2 + y^2 = 10$ and $xy = z$, where $z$ is one of the above roots, we obtain $x + y = {28 \over 10 - z}$, where $z$ can be any of the three roots. Now that we have $xy = z$ and $x + y = {28 \over 10 - z}$, for each of the three values of $z$ we may solve the quadratic equation $$a^2 - {28 \over 10 - z}a + z = 0$$ For each pair $\{r_1,r_2\}$ solving the above equation, $(x,y) = (r_1,r_2)$ and $(x,y) = (r_2,r_1)$ will potentially solve the original equation. I will leave it to you to test which pairs actually solve it. But note that for $z = 3$ we get solutions $(3,1)$ and $(1,3)$.

Answer 3

By Girard-Newton identities $$p_2=p_1e_1-2e_2$$ $$p_3=p_2e_1-p_1e_2+3e_3$$ and $p_2=10, p_3=28, e_3=0, p_1=e_1=x+y$ we have $$(e_1-4)(e_1^2+4e_1-14)=0.$$ So $x+y=4$ is the rational solution.

Answer 4

From the first equation,

$(x, y) = \sqrt{10} ( \cos \theta , \sin \theta ) $

Plug this into the second equation, then

$28 = 10 \sqrt{10} ( \cos^3 \theta + \sin^3 \theta ) $

Use the transformation $ z = \tan\left(\dfrac{\theta}{2}\right) $, then

$ \cos \theta = \dfrac{1 - z^2}{1 + z^2} $ and $ \sin \theta = \dfrac{2z}{1 + z^2} $

Using these, we get

$ K = \dfrac{ 28 }{10 \sqrt{10}} = \left( \dfrac{1 - z^2}{1 + z^2} \right)^3 + \left( \dfrac{2z}{1 + z^2} \right)^3 $

Multiplying through by $(1 + z^2)^3 $, this becomes

$ K (1 + z^2)^3 = (1 - z^2)^3 + (2 z)^3 $

Which is a sixth degree polynomial in $z$. Solving we get the $4$ real roots which are:

$ -0.13079, 0.162278, 0.720759, 1.300937 $

From these we apply the inverse transform to get $\theta$'s:

$\theta = 2 \tan^{-1} z $

then we compute $x$ and $y$ for each $\theta$.

$(x, y) = \sqrt{10} (\cos \theta , \sin \theta) $

The $(x,y)$ solutions are

$ (3.055911, -0.81327) $

$(3, 1) $

$ (1, 3) $

$ (-0.81327 , 3.055911 ) $

Answer 5

I tried to decompose the $x^3+y^3$, but this doesn't help

I think this means you factored the term. But this will help as the following shows.

$$\begin{align} x^2 + y^2 &= 10 \tag{1}\\ x^3 + y^3 &= 28 \tag{2} \end{align}$$ If we want to find integer solutions, we only have to check $x \in \{-3,-2,\ldots,3\}$ because of $(1)$ and we will find $(1.3),(3,1)$. I have no systematic way to find a solution but here the following works.

We factor $(2)$ to get

$$(x+y)(x^2-xy+y^2)=28\tag{3}$$ and now we square this to get $$(x^2+2xy+y^2)(x^2-xy+y^2)^2 =28^2 \tag{4}$$ Here we substitute $x^2+y^2$ by $10$ according to $(1)$ and get $$(2z+10)(10-z)^2=28^2\tag 5$$ if we set $$z=xy\tag 6$$ $(5)$ is a cubic equation and we know that one solution is $z_0=3\cdot 1$. Therefor $(z-3)$ is a factor of $(5)$ and we have to solve the quadratic equation $$z^2-12z-36=0\tag 7$$ and the solutions $z_1$ and $z_2$. From $(3)$ we get $$x+y=\frac{28}{10-z_k} \tag 8\\ xy=z_k$$ This finally gives us the qquadratic equation $$x(\frac {28}{10-z_k}-x)=z_k \tag{10}$$