<s>[INST] 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?
 [/INST] 
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.

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