Polynomial $p\in\mathbb Q[x,y]$ such that $p(e,\pi)=0$. Is there any polynomial $p\in\mathbb Q[x,y]$ that $p(e,\pi)=0$?
I was inspired by the fact that $\pi^4+\pi^5-e^6\approx 0$ and this lead me to the aforementioned question. Since there is no $p\in\mathbb Q[x]$ that $p(\pi),p(e)=0$ and the non-polynomial $f(e,\pi)=e^{i\pi}+1=0$. Is the statement in question a well-known result from somewhere? Any help would be appreciated, but please keep it as a (little as possible) hint so that I can study it further. Thank you in advance.
 A: In the comments already a few easy examples are mentioned. Just to list them in an answer here as well.

*

*You could take $p(x, y) = xy - 1$ and then pick any transcendental $x$ and set $y = 1/x$ (as suggested by Pedro Tamaroff).

*A similar strategy: take $p(x, y) = x - y^2$ and again pick any transcendental $x$ and now set $y = \sqrt{x}$.

You did also ask about a polynomial $p(x, y)$ such that $p(e, \pi) = 0$. This is still an open problem (as mentioned in the comments as well). It is conjectured that there is no such $p$. That is, that $e$ and $\pi$ are algebraically independent.
You might be interested in Schanuel's conjecture, which is the following much more general statement.

Let $z_1, \ldots, z_n \in \mathbb{C}$ be linearly independent over $\mathbb{Q}$ then the transcendence degree of $(z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n})$ (over $\mathbb{Q}$) is at least $n$.

So to see how this implies that $e$ and $\pi$ are algebraically independent we can consider $z_1 = 1$ and $z_2 = \pi i$. Then by Schanuel's conjecture we would have that $(\pi i, 1, -1, e)$ has transcendence degree at least 2. Since $1$ and $-1$ are algebraic we have that $\pi i$ and $e$ must be algebraically independent, and hence $\pi$ and $e$ must be algebraically independent.
The conjecture is much more general than that and is studied a lot in model theory. This is also mentioned in the MO post that leoli1 links to. One nice thing to highlight there is that Zilber proved that there is indeed an algebraically closed field $\mathbb{B}$ with an exponential function (i.e. a group homomorphism from the additive group of $\mathbb{B}$ to the multiplicative group of $\mathbb{B}$), which satisfies Schanuel's conjecture.
