Olympiad Math Problem About Simultaneous Equations Given the question:

*

*$x + y = 1$

*$x^2 + y^2 = 2$

*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substituting this into $2)$ we get a quadratic $2y^2 - 2y - 1 = 0$.
However, when we solve for $y$ we get two possible solutions. Hence how do we proceed from here? Why doesn't simultaneous equations work to solve the above problem?
 A: 
Why doesn't simultaneous equations work to solve the above problem?

It does work. You just need to carry the solution far enough.
To start, note that $x$ and $y$ are symmetric. In other words, you can replace $x \rightarrow y$ and $y \rightarrow x$ and you'd still be solving the same problem. Whatever value $x, y$ have, the same values for $y, x$ will also solve the problem. Hence you should expect there to be two possible solutions for both $x$ and $y$.
If you solve the quadratic, you find:
$y = 1/2(1+\sqrt{3})$ or $y = 1/2(1-\sqrt{3})$
Putting these solutions into $x + y = 1$, you get
if $y = 1/2(1+\sqrt{3})$ then $x = 1/2(1-\sqrt{3})$, and if $y = 1/2(1-\sqrt{3})$ then $x = 1/2(1+\sqrt{3})$.
The solutions are interchangeable, as expected from the discussion above about $x$ and $y$ being symmetric.
From here it's just a matter of putting these expressions into $x^5 + y^5$. You only need to do this calculation once, because $x$ and $y$ are symmetric. Taking the fifth power is tedious but straightforward with pen and paper, and trivial if you're allowed a calculator. The result is the answer.
(The other answers give more elegant ways to solve the problem.)
A: First note that we can get
$$2xy = (x+y)^2 - x^2-y^2 = 1 - 2 = -1$$
We can then apply binomial theorem again to obtain $x^3+y^3$ and $x^5+y^5$
$$
\begin{align*}
x^3 + y^3 &= (x+y)^3 - 3x^2y - 3xy^2 \\
&= 1 + \frac{3}{2}(x+y) = \frac{5}{2} \\
x^5 + y^5 &= (x+y)^5 - 5x^4y - 10x^3y^2 - 10x^2y^3 - 5xy^4 \\
&= 1 + \frac{5}{2}(x^3+y^3) - \frac{5}{2}(x+y) \\
&= 1 + \frac{25}{4} - \frac{5}{2} = \frac{19}{4}
\end{align*}
$$
A: 
We might also think of the first two equations as describing the intersections of a line with intercepts $ \ (0 \ , \ 1) \ \ $ and $ \ (1 \ , \ 0 ) \ $ and the circle of radius $ \ \sqrt2 \ $ centered on the origin.  These intersections are symmetric about the line $ \ y  \ = \ x \ \ . $
Since the axis intercepts are within the circle, the intersection points occur in the second and fourth quadrants, so their coordinates are $ \ (-X \ , \ Y) \ \ $ and $ \ (Y \ , \ -X) \ \ . \ $  The numerical result for $ \ x^5 + y^5 \ \ $ will thus be the same for both points.
We can solve the two equations simultaneously then as
$$ Y \ \ = \ \ 1 \ - \ X \ \ \Rightarrow \ \ Y^2 \ \ = \ \ (1 - X)^2 \ \ = \ \ 2 \ - \ X^2 \ \ \Rightarrow \ \ 2X^2 - 2X - 1 \ \ = \ \ 0 \ \ , $$
for which the solutions are $ \ X \ = \ \frac12  \pm \frac{\sqrt3}{2}  \ \ , \ $ corresponding to the fourth- and second-quadrant solutions.
From this, we obtain
$$  X^2 \ \ = \  \ X \ + \ \frac12 \ \ = \ \ 1 \pm \frac{\sqrt3}{2} \ \ , \ \ Y^2 \ \ = \ \ 2 \ - \ X^2 \ \ = \ \ 1 \ \mp \frac{\sqrt3}{2} $$
$ \Rightarrow \  \ \mathbf{X^5 \ + \ Y^5} \ \ = \ \ X·(X^2)^2 \ + \ Y·(Y^2)^2 $
$$ = \ \ \left( \frac12  \pm \frac{\sqrt3}{2} \right)·\left( 1 \pm \frac{\sqrt3}{2} \right)^2 \ + \ \left( \frac12  \mp \frac{\sqrt3}{2} \right)·\left( 1 \mp \frac{\sqrt3}{2} \right)^2 $$ $$ = \ \ \left( \frac12  \pm \frac{\sqrt3}{2} \right)·\left( \frac74 \pm  \sqrt3   \right)  \ + \ \left( \frac12  \mp \frac{\sqrt3}{2} \right)·\left( \frac74 \mp  \sqrt3  \right) $$
[all of the "middle terms" involving $ \ \sqrt3 \ $ cancel!]
$$ = \ \ \frac12·\frac74 \ + \ \frac32 \ + \ \frac12·\frac74 \ + \ \frac32 \ \mathbf{= \ \frac{19}{4} } \ \ . $$
The green curve in the graph represents this last equation.

A: HINT:
$$(x^2+y^2)\left((x+y)^3-3xy(x+y)\right)=\color{red}{(x^5+y^5)}+(xy)^2(x+y)$$
where, $xy=\frac{(x+y)^2-(x^2+y^2)}{2}$.
A: $$1=(x+y)^2= x^2+2xy+y^2 = 2+2xy$$$$ \implies xy = -1/2$$
$$4=(x^2+y^2)^2 =x^4+2x^2y^2 +y^4$$$$\implies x^4+y^4= 7/2$$
and $$2=(x+y)(x^2+y^2) = x^3+xy(x+y)+y^3$$ $$\implies x^3+y^3 = 5/2
$$
So
$$7/2=(x+y)(x^4+y^4) = x^5+y^5 + xy(x^3+y^3)$$ $$\implies x^5+y^5 =19/4$$
