Mathematica competitions in Croatia

a) Assume that $a=x+y$, $b=x^2+y^2$ and $c=x^4+y^4$ are integers. Then the numbers $a^2-b=2xy$ and $b^2-c=2x^2{y}^2$ are integers as well.

Suppose that $xy$ is not an integer. Then $xy=\frac{m}{2}$, where $m\in \mathbb{Z}$ is odd. But, then $2x^2{y}^2=\frac{m^2}{2}$ is not an integer. Contradiction! Hence, $xy$ is an integer.

By induction we can prove that $x^n+y^n$ is an integer for each $n\in \mathbb{N}$:

Basis of induction: numbers $x+y$ and $x^2+y^2$ are integers, by assumption.

Let's suppose that, for some integer $n>1$, the numbers $x^{n-1}+y^{n-1}$ and $x^n+y^n$ are integers.

Then $$\begin{array}{rl}{x}^{n+1}+y^{n+1}& =(x^n+y^n)(x+y)-(x^{n}y+x{y}^n)\\ & =(x^n+y^n)(x+y)-xy(x^{n-1}+y^{n-1}).\end{array}$$ Numbers $x+y$ and $xy$ are integers, as well as the numbers $x^{n-1}+y^{n-1}$ and $x^n+y^n$, so, we conclude that the number $x^{n+1}+y^{n+1}$ is an integer, too.

b) $x=\sqrt{2}$, $y=-\sqrt{2}$.

c) $x=\frac{1}{\sqrt{2}}$, $y=-\frac{1}{\sqrt{2}}$.