Is there a nice, short and elementary argument that the field extension $\mathbb{R}(X+Y)\subseteq\mathbb{R}(X,Y)$ is purely transcendental?
Obviously, $\mbox{tr deg}_{\mathbb{R}(X+Y)}\mathbb{R}(X,Y)\le1$, because $\mathbb{R}(X+Y)\subseteq\mathbb{R}(X+Y,Y)=\mathbb{R}(X,Y)$, so it is left to show that $Y$ is not algebraic over $\mathbb{R}(X+Y)$.
I don't see any nice proofs of this fact, only some brute force methods of summing degrees of powers of $Y$ in polynomials from $\mathbb{R}(X+Y)[\mathbb{X}]$.
Similar question concerns the transcendence degree of the extension $\mathbb{R}(X^2+Y^2)\subseteq\mathbb{R}(X,Y)$. This extension is not purely transcendal (an easy proof using automorphisms from Galois group). $X$ is algebraic over $\mathbb{R}(X^2)$, so again $\mbox{tr deg}_{\mathbb{R}(X^2+Y^2)}\mathbb{R}(X,Y)\le1$, because $\mathbb{R}(X^2+Y^2)\subseteq\mathbb{R}(X^2+Y^2,Y)=\mathbb{R}(X^2,Y)\subseteq\mathbb{R}(X,Y)$. But how to show that $Y$ is not algebraic over $\mathbb{R}(X^2+Y^2)$?
I don't know algebraic geometry, thus please don't use it in your answer.