# Diophantine equation without unique formula for solutions

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?

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We can always transform a parametrization into an equivalent but "different-looking" parametrization. – André Nicolas Jan 23 '14 at 15:08
Yes, my question was not clear... I hope it is now – Macadam Jan 23 '14 at 16:55

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.

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If I well understand for the equation $x^2+47y^2=z^3$ one has two parametrizations of solutions: $((p^3-3dpq^2)t^3)^2 + 47((3p^2q-dq^3)t^3)^2 = ((p^2+dq^2)t^2)^3$ and $(13u^3+30u^2v-42uv^2-18v^3)^2 + 47(u^3-6u^2v-6uv^2+2v^3)^2 = 2^3(3u^2+uv+4v^2)^3$ which give different solutions (ie there's at least one solution from (3) that is not cover by (2) and there is at least one solution from (2) that is not cover by (3)), but there's maybe a unique parametrization but we do not know it from yet. In this case we are not sure that (2)+(3) cover all solutions? – Macadam Jan 24 '14 at 9:09
Yes, you understood it correctly. But regarding your last sentence, I am of the opinion that (2)+(3) will not cover all solutions. In fact, since (1) is a non-homogeneous equation for $n>2$, I do not think there will be a unique polynomial parameterization at all that will solve it like the case $n=2$. – Tito Piezas III Jan 24 '14 at 18:11

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.

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