0
$\begingroup$

By saying $\forall\exists$ diophantine equations I mean sentences of the form: $\forall x\exists y\,[p(x,y) = 0]$ where $p$ is a polynomial on $x,y$, and both $x,y$ range over natural numbers.

I want to know the decidability of these sentences. First of all, we can rewrite the sentence into $\forall x\exists y\,[a_n(x)y^n+\cdots+a_1(x)y+a_0(x)]$. If $a_0(x)$ is a constant, then all possible values of $y$ must be divisor of $a_0(x)$. Assume $d_1,...,d_k$ are non-negative divisors of $a_0(x)$ and we can rewrite the problem into $\forall x\,[p(x,d_1)=0\lor \cdots\lor p(x,d_k) = 0]$ = $\forall x\,[(\Pi_{i=1}^k p(x,d_k)) = 0]$, which is trivially decidable by examining whether all coefficients of $x^n$ are zeros or not.

So what if $a_0(x)$ contains $x$? A reasonable guess is to assign $y$ with all polynomial divisor of $a_0(x)$. Is that sufficient? And how can we guarantee that the divisor of $a_0(x)$ must be non-negative?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.