Consider a special case of the Hilbert's 10th problem:


$f(\vec{x})=g(\vec{y})$, where $\vec{x}$ and $\vec{y}$ are **disjoint** ( i.e, the LHS and RHS do not have any common variables), moreover, $f$ and $g$ are polynomials with **positive** coefficients. 


The question is, with the restrictions, whether the undecidability still holds?


For example, the Pell equation $x^2=3y^2+1$ falls into this class and is difficult to solve. I am wondering whether we can obtain undecidability outright.


[**Important note**: the Hilbert 10th question has two versions, ie., whether there is a solution in natural numbers, or in integers. In general, they are equivalent, but in this particular case, they are not. The answer below shows that checking whether there is a solution in integers is undecidable, but it is still open for the case that the solution must be natural numbers.]



Undecidability of Diophantine equations with disjoint variables?
It is undecidable.


The only integral point on $x^3+x=y^2$ is $(0,0)$.


Let $F(\vec{y})=0$ be undecidable diophantine equation with
positive coefficients and not depending on $x$.


Take $f(x)=x^3+x$ and $g(\vec{y})=F^2$ leading to $x^3+x=F^2(\vec{y})$.


To get $F$ from $F'$ with negative coefficients use sum of squares
replacing each negative coefficient $c\_i$ with variable $v\_i$ and
add the square $(v\_i + c\_i)^2$.


Even simpler, the integral solutions of $x^2=1+d^2 y^2$ are $(\pm 1,0)$.


