Let $C\subset\Bbb{R}^2$ be an algebraic curve defined by an irreducible polynomial $P$.
Take $Q\in\Bbb{R}[x,y]$ an another polynomial such that $Q(x,y)=0$ on a set $A\subset C$ such that $A$ admit a limit point.
Does it follow that $P$ divides $Q\;?$
I do not know much about algebraic curves. I think that I have seen this property during a seminar. But it's been a while.
Is it possible to give an elementary proof?
On
It's actually not that hard to give an elementary proof of what Mohan claims. Slightly more generally, we have the following.
Proposition. Let $P$, $Q \in \mathbb R[x,y]$ be polynomials. If $P$, $Q$ are coprime, then the curves $C(P)$ and $C(Q)$ have only finitely many points in common.
Lemma. There exist $A$, $B \in \mathbb R[x,y]$ such that $AP+BQ \in \mathbb R[x]$.
Proof: View $P$, $Q$ as elements of $\mathbb R(x)[y]$, that is, as polynomials in $y$ over rational functions in $x$. Since $\mathbb R(x)$ is a field, $\mathbb R(x)[y]$ is a Euclidean domain (using polynomial division). By coprimality of $P$, $Q$ we can therefore find $A_0$, $B_0 \in \mathbb R(x)[y]$ such that $A_0P + B_0Q = 1$. Letting $D \in \mathbb R[x]$ be a common denominator of $A_0$ and $B_0$, and setting $A=DA_0$, $B=DB_0 \in \mathbb R[x,y]$, we have $AP+BQ=D \in \mathbb R[x]$.
Proof of Proposition: Let $A$, $B \in R[x,y]$ with $AP + BQ = D \in \mathbb R[x]$. If $(x_0,y_0)$ is a common zero of $P$ and $Q$, then $D(x_0)=0$. Since $D$ is a univariate polynomial, there are only finitely many possibilities for $x_0$. For each such $x_0$, we have that $P(x_0,y)$ and $Q(x_0,y)$ are univariate polynomials in $y$.
If $P(x_0,y)=Q(x_0,y) = 0$, then $(x-x_0)$ divides both $P$ and $Q$, in contradiction to our assumptions. (To see that $(x-x_0)$ divides $P$, write $P=\sum_{i=0}^n P_i(x) y^i$, and note that $P_i(x_0)=0$ implies that $(x-x_0)$ divides each $P_i$.) Thus at least one of $P(x_0,y)$ and $Q(x_0,y)$ is not identically zero, say $P(x_0,y) \ne 0$. Since $P(x_0,y)$ is a nonzero univariate polynomial in $y$, it has only finitely many zeroes.
To recap, we have shown that for common zeroes $(x_0,y_0)$, there are only finitely many possibities for $x_0$, and that for each fixed such $x_0$, there are only finitely many possibilties for $y_0$. This proves the claim. $\square$
The full Bézout's Theorem, which we haven't proved here, would also give you a bound on the number of common points in terms of the degrees of the polynomials.
After general change of variables, you may assume $P(x,y)=y^n+a_1(x)y^{n-1}+\cdots+a_n(x)$, where $a_i(x)\in\mathbb{R}[x]=A$. If $B$ denotes $\mathbb{R}[x,y]/P$, it is a free module over $A$ generated by $1,y,\ldots, y^{n-1}$ and thus multiplication by $Q:B\to B$ can be thought of as an $n\times n$ matrix over $A$. Since $P$ is irreducible, if $P$ does not divide $Q$, then this map is injective and thus $g(x)=\det Q\neq 0$. It is immediate that if $(a,b)$ is a common zero of $P,Q$, then $g(a)=0$. Thus, the first co-ordinates of the common zeros are at most finite, being zeros of $g$. Since $P$ is monic in $y$, it follows that for a fixed $x$, there are only finitely many $y$'s satisfying the equation and thus the common zeros of $P,Q$ is finite.