Real numbers $x$ , $y$ and $z$ are such that $x+y=2$ and $xy-z^2=1$ . Find $x^5+y^5+z^5$ .

The answer is 2.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

$(x-y)^2=(x+y)^2-4xy=2^2-4(z^2+1)$ $\implies (x-y)^2=-4z^2$ Since $x,y$ are real, the LHS is non-negative, so the RHS must be non-negative. This is possible only when $z=0$ and $x=y$ . Solving the equations, we get $x=1, y=1$ .

Therefore, $\boxed{x^5+y^5+z^5=1^5+1^5+0^5=2}$