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