Diophantine equation without unique formula for solutions

Question

Every one know solutions of the Diophatine equation $x^2+y^2=z^2$ which are given by formula $x=t(a^2-b^2)$, $y=t(2ab)$ and $z=t(a^2+b^2)$. In this exemple one proove that all the solutions are in this form.

My question is (Question 0): does all the solvable Diophantine equations have a unique complete description?

I think the word "description" is ambiguous. We could for exemple in a first approach require having description in polynomial form function of a finit number of parameters. So the more precise question could be:

Question 1: Is there a Diophantine equation (which has solution) for which we can proove that one parametrization in polynomial form is not enough to describe all solutions

A easier variant should be:

Question 2: Is there a Diophantine equation (which has solutions) for which we don't have yet find a unique polynomial expression for the solutions?

Answer 1

It is even worse than that. Matiyasevitch's solution to Hilbert's tenth problem showed there are Diophantine equations which we can't decide whether they have solutions or not.

Answer 2

Even an equation as simple as,

$$x^2+dy^2 = z^n\tag{1}$$

already becomes tricky once we go $n>2$. We can solve (1) for any positive integer $n$ as,

$$(x+y\sqrt{-d})\,(x-y\sqrt{-d})=(p+q\sqrt{-d})^n\,(p-q\sqrt{-d})^n$$

equating factors, and solving for $x,y$. This method (after scaling) gives the complete solution for $n=2$. For $n=3$, this yields,

$$((p^3-3dpq^2)t^3)^2 + d((3p^2q-dq^3)t^3)^2 = ((p^2+dq^2)t^2)^3\tag{2}$$

but is no longer complete. For example, for $d=47$, Pepin found there is no rational $p,q,t$ that corresponds to the solution,

$$(13u^3+30u^2v-42uv^2-18v^3)^2 + 47(u^3-6u^2v-6uv^2+2v^3)^2 = 2^3(3u^2+uv+4v^2)^3\tag{3}$$

Thus (1) for $n>2$ is an example for your Question 2.