# Lecture 13

## Hilbert's 10th Problem

Definition A polynomial is a sum of terms where each term is a coefficient times a product of some variables.

In this lecture, all polynomials have integer coefficients.

H10v1: Is there an algorithm where for any input that is an expression involving polynomials, the algorithm determines wheter there is any assignment of integer values to the variables that satisfy the equation?

H10v2: Is there any algorithm that given a polynomial $$P$$ on $$n$$ variables, determines if there exists $$x\in\mathbb{Z}^n$$ such that $$P(x)=0$$?

Theorem: H10v1=H10v2

Proof: $$P(x)=Q(x)\Longleftrightarrow R(x) = 0$$ where $$R=P-Q$$

H10v3: $$x\in\mathbb{N}^n$$

Theorem: H10v1=H10v2

Proof: Let $$A_3$$ be an algorithm that solves H10v3.

Let $$P$$ be the input of H10v2. Construct

$Q(x_1,y_1,x_2,y_2,...,x_n,y_n)=P(x_1-y_1,...,x_n-y_n)$

Call $$A_3$$ on $$Q$$. If $$A_3$$ accepts, then $$\exists x\in\mathbb{Z}^n$$ such that $$P(x)=0$$

If there exists $$z\in\mathbb{Z}^n$$, I can always set $$x_i=a_i+1,y_i=1$$ when $$z_i\geq 0$$, $$y_i=z_i+1,x_i=1$$ when $$z_i<0$$ and have $$Q(x_1,y_1,...,x_n,y_n)=0$$ so $$A_3$$ accepts.

The other direction:

Lagrande's Theorem: Every positive integer can be expressed as the sum of $$4$$ squares of integers.

H10v4: Determine whetre the language $$L_{H10}=\{<P>:P$$ is a polynomail s.t. $$\exists x\in\mathbb{N}^n$$ for which $$P(x)=0\}$$ is decidable.

## Diophantine Sets

Definition: The $$S\subseteq\mathbb{N}$$ is a Diophantine set if there is a polynomial $$P_s$$ such that $S=\{a\in\mathbb{N}:\exists y_2,...,y_n\in\mathbb{N}^n\quad s.t.\quad P_s(a,y_2,...,y_n)=0\}$

Claim: $$S=\mathbb{N}$$ is Diophantine

Proof: $$P_s(x,y)=x-y$$

Claim: $$S=$$even numbers

Proof: $$P_s(x,y)=x-2y$$

Claim: $$S=$$not powers of $$2$$

Proof: $$P_s(x,y,z)=x-(2y+1)z$$

Claim: $$S=$$composite numbers

Proof: $$P_s(x,y,z)=x-(y+1)(z+1)$$

Definition: The language corresponding to $$S\in\mathbb{N}$$ is $$L_S=\{<x>:x\in S\}$$

Theorem Is there is a Diophantine set $$S$$ such that $$L_S$$ is undecidable, then $$L_{H10}$$ is also undecidable.

Proof Let $$S$$ be a Diophantine set s.t. $$L_s$$ is undecidable. Let $$P_s$$ be the associate polynomial.

Assume that $$L_{H10}$$ is decidable. Let $$M$$ be a TM that decides $$L_{H10}$$ Construct $$M'$$ that on unput $$<x>$$

1. Constructs $$Q(y_2,...,y_n)=P(x,y_2,...,y_n)$$
2. Call $$M$$ on $$<Q>$$
3. Accept if $$M$$ accepts, reject if $$M$$ rejects.

## Davis' Conjecture

DPRM Theorem: Every set $$S\subseteq\mathbb{N}$$ such that $$L_s$$ is recognizable is Diophantine

Theorem If DC holds, then there is a polynomial $$P$$ whose set of positive values that if takes is exactly the set of prime numbers.