TITLE: Define $\mathbb{N}$ in the ring $\mathbb{Z}$ without Lagrange's theorem
QUESTION [30 upvotes]: It is well-known that the set of nonnegative integers $\mathbb{N}$ is definable in the ring of integers $\mathbb{Z}$. Indeed, by Lagrange's four squares theorem we have $\mathbb{N} = \{n \in \mathbb{Z} : \varphi(n)\}$, where $\varphi$ is the formula
$$\varphi(x) := \exists a\, \exists b\, \exists c\, \exists d \; x = a^2 + b^2 + c^2 + d^2$$
However, Lagrange's theorem is not so trivial, so I wonder:
Is there a more elementary and self-contained proof of the definability of $\mathbb{N}$ in the ring of integers?
Thank you.

REPLY [35 votes]: Raphael Robinson, in a paper entitled Arithmetical Definitions in the Ring of Integers, gives a definition of $\mathbb{N}$ in $\mathbb{Z}$ using only two existential quantifiers. He does not use Lagrange's Theorem, but he does use the fact that the equation $y^2-az^2=1$ has infinitely many solutions $y$ and $z$ whenever $a$ is a positive nonsquare integer. Whether this is "easier" than Lagrange's Theorem is debatable.
Robinson proves that $x\in \mathbb{N}$ if and only if
$$\exists y\exists z \left(x=y^2\vee (y^2=1+xz^2\wedge y^3\ne y)\right).$$
He also proves that no definition using just one existential quantifier is possible.