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?


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.

