Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$.

Question

My attempt:

Let $x, y$ be roots of $$t^2-5t+p=0$$ I got $$p=\frac{5 \pm \sqrt{-431}}{2}$$ using Vieta's relations. Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$. But I got two very hideous complex values of x and y. So I think my answer may be wrong.

The values I got are $$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or $$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$

Please post your own solutions too if you have any different solutions.

Answer 1

Hint: Substituting $y=5-x$ the equation $x^5+y^5=275$ is equivalent to $$ (x^2 - 5x + 19)(x - 2)(x - 3)=0 $$

Answer 2

Observing that the pairs $(2,3)$ and $(3,2)$ are solutions and substituting $y=5-x$ from the second equation into the first one one obtains: $$ (x^2 - 5x + 19)(x - 2)(x - 3)=0. $$

Can you take it from here?

Answer 3

$$x^3+y^3=(x+y)^3-3xy(x+y)=5^3-3xy(5)$$

$$x^2+y^2=(x+y)^2-2xy=5^2-2xy$$

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

$$\iff(125-15xy)(25-2xy)=275+5x^2y^2$$

$$\iff(xy)^2-25xy+114=0$$

For the rest, follow How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?