With linear algebra, prove that all polynomial parametric curves can fulfill a polynomial Cartesian equation

This problem in my linear algebra class is intended to demonstrate that all polynomial parametric curves (in $$\mathbb{R^2}$$) can fulfill a polynomial Cartesian equation. First, we let $$x = p(t)$$ and $$y = q(t)$$ where $$p,q \in \mathcal{P}\mathbb{(R)}$$ are fixed polynomials dependent on the variable $$t$$.

Part (a) wants us to find a function, $$L$$, that takes nonnegative integers $$(m,n)$$ and returns a nonnegative integer $$L(m,n)$$ so that if $$0 \leq i \leq m$$ and $$0 \leq j\leq n$$, then $$x^iy^j = p(t)^iq(t)^j \in \mathcal{P}_{L(m,n)} (\mathbb{R})$$. This choice of $$L$$ also depends on $$p$$ and $$q$$ in a way.

Part (b) wants us to choose $$m$$ and $$n$$ from above so that the number of monomials $$x^iy^j$$ with $$0 \leq i \leq m$$ and $$0\leq j \leq n$$ exceeds the dimension of $$\mathcal{P}_{L(m,n)}(\mathbb{R})$$.

Part (c) wants us to show that there exists a nonzero two-variable polynomial $$f$$ in which $$f(x,y) = f(p(t),q(t)) = 0$$. That is, $$f(x,y)$$ in the form $$f(x,y) = \sum_{i,j} c_{i,j}x^iy^j (c_{i,j} \in \mathbb{R})$$ where the sum is finite.

My attempt at a solution:

First, I called the degree of $$p(t)= a$$, and the degree of $$q(t) = b$$. For part (a), I created a function which would, given $$m,n$$, return a number that would mark the highest possible order of polynomial that could result from $$p(t)^iq(t)^j$$. So, I let $$L(m,n) = am + bn$$ which comes from exponentiating polynomials $$p(t)$$ and $$q(t)$$ and adding exponents for the final polynomial product. All polynomials of the form $$p(t)^iq(t)^j$$ are now in $$\mathcal{P}_{L(m,n)} (\mathbb{R})$$.

For part (b), I deduced that the number of possible monomials $$x^iy^j$$ is $$(m+1)(n+1)$$ since $$0 \leq i\leq m$$ and $$0 \leq j \leq n$$. In addition, dim$$\mathcal{P}_{L(m,n)}(\mathbb{R}) = L(m,n) + 1$$, or, for my function, $$am + bn + 1$$. So, I have to come up with $$m,n$$ such that $$(m+1)(n+1) > am + bn + 1$$, but this boils down to $$mn + m + n > am + bn$$. I tried various combinations of letting $$m,n$$ be $$0, 1, a,$$ and $$b$$, but could not satisfy the inequality.

Is there something I'm misinterpreting with the problem, or is my function a poor choice in satisfying the conditions in the problem? What type of function would satisfy the conditions?

I was not able to move on to part (c), but I'm not sure how, once (b) is satisfied, one could deduce the conclusion that part(c) wants us to show. What does the result of part(b) indicate? Does it have to do with the number of polynomials in the basis of $$\mathcal{P}_{L(m,n)}(\mathbb{R})$$ exceeding the dimension and creating a contradiction or something? Why is it important that $$f(x,y) = 0$$?

Let $$c=\max\{a,b\}$$ and let $$m>2c, n>2c$$.
Then $$mn > 2c\max\{m,n\} \geq cm + cn \geq am +bn$$ so $$mn+m+n$$ as well.
Then for part c) the idea is that $$(x^iy^j)_{i,j}$$ is family that lives in a vector space but has size larger than the dimension : therefore it must be linearly dependent.
$$f(x,y)=0$$ just tells you that $$x$$ and $$y$$ “fulfill a polynomial equation“ which is what you wanted in the first place