Proving that all integers are even or odd I know that $\mathbb{Z}$ is a group under addition with a multiplication defined. I have just the definition of even and odd integers: $n$ is even if $n = 2k$ for some integer $k$ and $n$ is odd if $n = 2k+1$ for some integer $k$.

Using just this I am wondering how to prove that all integers are either even or odd. That is, how can I prove that given an integer $n$, $n$ must be even or odd?

My problem with this is that it seems so simple. I know that one can divide an integer by $2$ and the remainder will be $0$ or $1$. Using this, it is clear that the even and odd integers make up everything. But how can I prove it without using this fact about remainders and such? I guess one could also use facts about prime numbers, but I am looking for a proof that just uses the definition of odd and even.
If you really want to avoid using general facts about division with remainder, you can use mathematical induction on $n$ to prove it for nonnegative integers:

Base case: $0$ is even because $0=2\cdot 0$.

Induction step: Assume that $n$ is odd or even; then we must prove that $n+1$ is also either odd or even.

First subcase: $n$ is even, so $n=2k$ for some $k$. Then $n+1=2k+1$ and so it is by definition odd.

Second subcase: $n$ is odd, so $n=2k+1$ for some $k$. Then $n+1=2k+1+1=2k+2=2(k+1)$, and so $n+1$ is even.

That completes the induction proof, and now we just need to know that negative integers are also all either odd or even. But if $n$ is negative, then $-n$ is positive. One easily sees that if $-n$ is even, then $n$ is even too, and if $-n$ is odd, then $n$ is odd too.



Whether this satisfies your requirements is a bit debatable, because the induction part of the proof is essentially the same as how you prove that division with remainder works in general.

However, you CANNOT prove that every integer is either odd or even solely from the fact that the integers form "a group under addition with a multiplication defined", that is a commutative ring with unit. The reason for this is that there are some rings where not all elements are either odd or even -- such as $\mathbb Z[X]$, the ring of polynomials with integer coefficients. Here the polynomial $X+1$ is neither twice something, nor twice something plus one.
If you know or believe that every integer can be expressed in decimal, then this is easy: A number is even or odd according to whether the units digit in its decimal representation is even or odd. This is because $10$ is even and every single-digit number is either even or odd.

Of course, proving that every integer can be expressed in decimal (or any base) requires induction.

And if you go that route, its nicer to prove directly that every integer can be expressed in binary and the look at the units digit, which is $0$ or $1$.