solve the quintic non-linear system

Question

$x^5+y^5=33$ and $x+y=3$

I tried a variety of things here, all too ugly to post. the most promising was doing long division $(x^5+y^5)/(x+y), $but it still looked kind of ugly and not easily factorable. I am looking to solve this with simple high school level algebra techniques. Can someone give me a hint? Much appreciated.

Answer 1

From the second $y=3-x$. Plug in the first equation and expand $$x^5+y^5=x^5+(3-x)^5=15 x^4-90 x^3+270 x^2-405 x+243$$ making the equation to be $$15 x^4-90 x^3+270 x^2-405 x+210=0$$ Divide all terms by $15$ to get $$x^4-6 x^3+18 x^2-27 x+14=0$$ which, by inspection, shows two simple roots.

Just continue.

Answer 2

The first thing I try is plug some random small integers to the equations and see whether they work. I already get two "obvious" solutions $(x,y) = (2,1)$ and $(1,2)$.

To see this exhaust all solutions, introduce a parameter $u$ such that $x = \frac{3+u}{2}$ and $y = \frac{3-u}{2}$.
In terms of $u$, we have

$$\begin{align} & x^5 + y^5 = 33\\ \iff & (3+u)^5 + (3-u)^5 = 33\cdot 2^5\\ \iff & 2(3^5 + 10\cdot 3^3 u^2 + 5\cdot 3 u^4) = 33\cdot 2^5\\ \iff & 10\cdot 3^2 u^2 + 5 u^4 = \frac13\left(\frac12\cdot 33\cdot 2^5 - 3^5\right) = 11\cdot16 - 81 = 85\\ \iff & u^4 + 18u^2 -19 = 0\\ \iff & (u^2-1)(u^2+19) = 0\\ \implies & u = \pm 1\end{align}$$ This means the equations have two and only two real solutions: $(x,y) = (2,1)$ and $(1,2)$.

Answer 3

Given $x+y=3$, raise to the fifth power:

$x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5=243$

Subtract $x^5+y^5=33$ from this and factor the remaining terms:

$5xy(x^3+2x^2y+2xy^2+y^3)=210$

$xy(x^3+2x^2y+2xy^2+y^3)=42$

We then have

$x^3+2x^2y+2xy^2+y^3=(x^3+3x^2y+3xy^2+y^3)-xy(x+y)=(x+y)^3-xy(x+y)=27-3xy,$

using the linear equation $x+y=3$. So

$xy(27-3xy)=42, xy(9-xy)=14$

$(xy)^2-9(xy)+14=0$

Thus $x+y=3$ and $xy\in\{2,7\}$. For each possible product we form a quadratic equation whose roots will give a pair of $x,y$ values. With $xy=2$ we have

$w^2-3w+2=0, w=x\text{ or }y\implies \{x,y\}=\{1,2\}$

With $xy=7$:

$w^2-3w+7=0, w=x\text{ or }y\implies \{x,y\}=\{(3/2)+\sqrt{19/4}i,(3/2)-\sqrt{19/4}i\}$

The quadratic equations are factors of the quartic found in Claude Lebovici's answer.