Your proof hinges on the not-directly-stated words "so, if x were 2, we would reach an impossibility" where impossibility == contradiction.

They may be equivalent, but not if we count proof by contradiction as special.

In response to edit: No, it doesn't. Having proven that x contains an even number of 2s as factors, it follows that x is not two. 2 does not have an even number of 2s as factors, x does, therefore they are different. No contradiction required.

The key point here is that you are partitioning the possibilities in a binary fashion: either x has even number of prime factors or it does not. You've shown that it must not have an even number of prime factors. You haven't shown that it isn't 2. You implied that if it were true, that would contradict what you have shown.

How you go from "X has an even number of prime factors" to "X can't be 2" is important.

You're right that they are "essentially" the same proof, but excluding contradictions is weird, a bit like excluding the letter e. Rephrasing a proof to not use it may be possible in some cases.

This is where you are implicitly employing proof by contradiction. Why do you take two cases and not three? Or four? What assumption allows you to restrict your conclusions like that?

Probably the Law of the Excluded Middle: (P or ~P). (This is equivalent to ~~p = p.)

Once you assume or derive either of those, you have validated proof by contradiction.

Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof.

When you break it down to the logical axioms, you can't use cases like that. Try it yourself. You get something like this:

Let A: (k > 0)

B: (k = 0)

k is a whole number, therefore A v B.

A -> X !=2

B -> X !=2

Okay, from here, you need to deduce X != 2. Traditionally we do this by applying proof by cases, but as I showed in my other comment, this requires the law of the excluded middle.

I don't see any other propositional logic axioms we can apply to achieve the same result.

I think that's because you don't know A v B. But in my case you can directly prove A v B.

The form is (A v B, A -> C, B -> C) -> C.

https://en.wikipedia.org/wiki/Proof_by_exhaustion says that this is called a direct proof.

Here is a more rigorous form of what you've said there:

    Let X be an integer with a prime factorization in the form 2^(2k) * p_1 * p_2 ... * p_n where k is a positive integer and p_i for any i is a prime not equal to 2.

    Proposition: X is not 2

    Proof:

    Assume X is 2.

    The prime factorization of X is:

    X = 2

    By the definition of X, we have:

    2 = 2^(2k) * p_1 * p_2 ... * p_n

    Since we defined k to be a positive integer, we have

    min(2^(2k)) = 2^(2(min(k)) = 2^(2*1) = 4

    So we have:

    2^(2k) >= 4

    So we have:

    2^(2k) > 2

    Since p_i for all i are all positive integers, we have:

    2 < 2^(2k) * p_1 * p_2 ... * p_n

    So we have:

    2 != 2^(2k) * p_1 * p_2 ... * p_n

    Thus:

    2 != X

    And we have arrived at a contradiction.

Consider k. Either it is 0, or greater. Consider 0. In that case, X is not divisible by 2, hence cannot be 2.

Now consider k positive. (My original proof didn't show k wa positive, so I needed to deal seperately.)

Use your proof to show that 2^(2k) > 2, and then 2!=X. As far as I can tell you never actually used the assumption that 2=x, so that can be deleted from the proof and it remains valid.