$p(x),q(x)$ be real polynomials with degree $\leq n (n> 1)$, both are non constants. How to find a $F(x,y)\neq 0$ such that $F(p(t),q(t))=0, \forall\ t\in\Bbb R$, where $F\neq 0$ is real polynomial with degree $\leq 2n-2$.
By [Two polynomial problem the last comment, such a $F$ with degree $2(2n-1)$ can be found. However, the restriction $\leq 2n-2$ could be reached?
As pointed out in Two polynomial problem, we can choose $F(x, y)$ as the resultant of $p(t) - x$ and $q(t) - y$ with respect to $t$.
In order to estimate the degree of $F$ we write the resultant as a Sylvester matrix: With $$ p(t) = a_m t^m + \ldots + a_0 \\ q(t) = b_n t^n + \ldots + b_0 $$ we have $$ F(x, y) = \operatorname{Res}(p - x, q - y) = \left|\begin{array}{ccc:ccc} a_m & & 0 & b_n & & 0 \\ \vdots & \ddots & &\vdots & \ddots & \\ a_1 & & a_m & b_1 & & b_n \\ a_0 - x & \ddots &\vdots & b_0 - y & \ddots & \vdots \\ & \ddots & a_1 & & \ddots & b_1 \\ 0 & & a_0 - x & 0 & & b_0 - y \end{array}\right| $$ where the left block of the determinant has $n$ columns, and the right block has $m$ columns.
In each term of the Leibniz expansion of this determinant, $x$ and $y$ contributions come only from the last $\max(m, n)$ columns, and are of the form $$ x^k y^l \text{ with } k+l \le \max(m, n) \, . $$
It follows that $\deg F \le \max(m, n) = \max(\deg p, \deg q)$.
In particular, if both $p$ and $q$ have a degree $\le n$ then $\deg F \le n$, and that is $\le 2n-2$ if $n > 1$.
