symmetric polynomial recursion to solve the system, $x^5+y^5=33$, $x+y=3$

Question

I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 \text{ , } x+y=3$$

In the text they said to denote $\sigma_1=x+y$ and $\sigma_2=xy$, and to use recursion. I understand how the formula below works $$S_5=x^5+y^5=\sigma_1 S_4-\sigma_2 S_3$$

But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?

Thanks

Answer 1

Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.

\begin{align} x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4) \\ c&= a(x^4+y^4-x^2xy+(xy)^2-xyy^2) \\ c&= x^4+y^4-b(x^2+y^2)+b^2 ,\\ x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2 ,\\ c&= a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2) = a(x^2+y^2)(x^2+y^2-b)-b^2) ,\\ x^2+y^2&=(x+y)^2-2xy=a^2-2b ,\\ c&= a(a^2-2b)(a^2-2b-b)-ab^2 , \end{align}

thus we have a quadratic equation in terms of $b$:

\begin{align} 5ab^2-5a^3b-c+a^5&=0 , \end{align}

\begin{align} b&=\frac{(5a^3\pm\sqrt{5a^6+20ac}}{10a} ,\\ b_1&=2 ,\\ b_2&=7 . \end{align}

So, we need to consider two cases: $xy=2$ and $xy=7$. In both cases we also have $x+y=3$.

Case 1 $xy=2$.

The quadratic equation with roots $x,y$ is \begin{align} t^2-3t+2&=0 ,\\ x,y&=1,2 . \end{align}

Indeed, $1^5+2^5=33$, $1+2=3$, so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.

Case 2 $xy=7$.

The quadratic equation with roots $x,y$ is \begin{align} t^2-3t+7&=0 ,\\ x,y&=\tfrac32\pm\tfrac{\sqrt{19}}2\cdot i , \end{align}

so in Case 2 we have two complex conjugate solutions, \begin{align} x&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i ,\\ y&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i \end{align} and \begin{align} x&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i ,\\ y&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i . \end{align}

Answer 2

Also, we can use the homogenization.

Let $y=tx$.

Thus, $$t^5+1=\frac{33(t+1)^5}{243}$$ or $$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.

Answer 3

Is there a shorter way to do the recursion to solve this specific problem?`

If you don't insist on recursion, the shorter way in this case is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:

$$ \begin{align} 0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \\ &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \\ &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7) \end{align} $$