Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. (http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic).

MA has arbitrarily large finite models based on modular arithmetic. All finite models of MA have either an even or odd number of elements. I call a model of MA "even" if it satisfies both of these two sentences:

E1) $\exists x(x \ne 0 \land x+x = 0)$

E2) $\forall x(x+x \ne S0)$

A model of MA is odd if it satisfies both of:

O1) $\forall x(x = 0 \lor x+x \ne 0)$

O2) $\exists x(x+x = S0)$

We can use compactness to prove MA has infinite "even" size models by adding the even definitions above as axioms. We can similarly prove there are infinite "odd" size models of MA. Some infinite sets, like the integers, are neither even nor odd. The integers are not the basis for a model of MA. For example, the four square theorem (every number is the sum of four squares) is a theorem of both MA and PA. The four square theorem is false in the integers. It has been conjectured the complex numbers are a basis for a model of MA. If so, the complex numbers would be an "odd" model of MA.

My question is whether every model of MA must be exclusively even or exclusively odd? Is this statement a theorem of MA?

$$\exists x(x \ne 0 \land x+x = 0) \ \overline{\vee}\ \exists x(x+x = S0)$$

I asked this question on stack exchange and got no answer.

https://math.stackexchange.com/questions/214018/even-xor-odd-infinities

[*The following was merged from an answer - ed.*]

Ashutosh's proof can be written as:

$\exists x\exists y( (x+x=0 \land y+y=1) \implies (x=0) )$

This answers my question when $\exists x(x+x=1)$ is true but it says nothing about when $\forall x(x+x \ne 1)$ is true. Emil and others have stated any algebraically closed field is a model of MA. Ashutosh's proof shows any algebraically closed field is odd because $\exists x(x+x=1)$ is true.

I want to accept Ben Crowell answer, but I have some reservations. The proof starts by showing how any model of MA can be expanded into a model of PA. I have made similar arguments and always assumed it would be easy to prove. My conjecture is true of all finite models of MA so we only need consider infinite models. MA is omega inconsistent and any infinite model must have non-standard elements. Tennebaum's theorem says addition is not recursive in non-standard models of PA. Can addition actually be recursive in $A$, the model of PA he constructs? It looks like he is assuming we can add non-standard numbers from the model of MA. I also wonder if he is assuming $I$ is a standard model of PA. I don't think it makes any difference, but it might.

Obo's proof is much simpler and similar to a proof I came up with. My proof had the same error as his. I think it is fixable. In the case where we have $S(y+y)=p$ we need to also prove $y \ne p$. $y \ne p$ can be true only in models with three or more elements.

This isn't a discussion group so I won't go into detail why I don't think the complex numbers are a model of MA. I don't think MA has any infinite models. I will point out MA proves a lot of interesting things about odd models. In an odd model the sum of all elements is 0. This can't be stated in first order. I think if we have a successor function defined on the complex numbers we can use it to order the reals. Just ignore numbers with a non-zero imaginary component.

I want to retract my statement that the Lagrange's four square theorem is a theorem of MA. I based my claim on Andrew Boucher's paper on General Arithmetic (GA). Boucher shows GA proves the four square theorem. I thought GA was a weak sub-theory of MA because GA has much weaker successor axioms. Rereading the paper I see Boucher says he is using 2nd order induction. He also says successor is second order.

area model of MA. The induction schema is valid in $\mathbb C$ because it is a minimal structure: any definable (with parameters) subset of $\mathbb C$ is finite or cofinite. If $\mathbb C\models\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))$ (where $\phi$ may have parameters not shown), let $X$ be the set defined by $\phi$. By the assumption, $X$ is infinite, hence its complement is finite. Since the complement is closed under subtracting $1$, this can only happen if it is empty. $\endgroup$ – Emil Jeřábek supports Monica Jan 21 '13 at 12:55