How can all of them be irrational ?? 

Assume that $\{x,y,x^2,y^2,xy\}$ are all irrational.
Can it be true that all of $\{x-y,x+y,x^2-y^2,x^2+y^2\}$ are irrational?
Details: $|x|\ne|y|$ and $x,y\in\mathbb R$.

In the question above I took $x$ to be $\sqrt{2} + 1$ and $y$ to be $\sqrt{2} - 1$ and thus both are irrational and their squares which are $x^2$ and $y^2$ are also irrational, so I get $x - y = (\sqrt{2} + 1) - ( \sqrt{2} - 1) = 2$ and $x^2 + y^2 = (\sqrt{2} + 1)^2 + ( \sqrt{2} - 1)^2 = 6$ to be rational. 
But in the answer above it says none could be rational !! So, do I digress somewhere in my reasoning ?

We need to show that, if any two of
$\{x-y,\; x+y,\; x^2-y^2,\; x^2 + y^2\}$
are rational then that at least one of 
$\{x,\; y,\; x^2,\; y^2,\; xy \}$
is rational. Which means that the answer to the question is no.
There are $6$ ways to choose two objects out of $4$.
Case 1, 2, and 3: Any two of $\{x-y,\; x+y,\; x^2 - y^2\}$ are rational.
Since $(x-y)(x+y) = x^2 - y^2$, then if any two are rational, then all three are rational. It follows then, that $\dfrac{(x-y)+(x+y)}2 = x$ is rational in all three cases.
Case 4: $x-y$ and $x^2 + y^2$ are rational
Then $\dfrac{(x^2 + y^2) - (x-y)^2}2 = xy$ is rational.
Case 5: $x+y$ and $x^2 + y^2$ are rational
Then $\dfrac{(x+y)^2 - (x^2 + y^2)}2 = xy$ is rational.
Case 6: $x^2-y^2$ and $x^2 + y^2$ are rational
Then $\dfrac{(x^2-y^2) + (x^2 + y^2)}2 = x^2$ is rational.
