"Every even whole number greater than 2 is the sum of two prime numbers."
$2n = p_1 + p_2\hspace{1cm} ,n>1 \\ 2n = [(p_1-1) +1] + [(p_2-1) +1] \\ 2n = (p_1-1) + (p_2-1) + 2 \\ 0 \equiv 0 \mod{(2)} $
Since $n>1$, then every prime number we are considering is odd. Therefore, both $p_1$ and $p_2$ can't be 2. Then $p_1$ and $p_2$ are 1 away from an even number; subtract 1 and add 1 for both $p_1$ and $p_2$. You can also choose to add 1 and subtract 1. Then modular 2 and we get a equivalence relation.
Is this a legit proof?
Edit: as I was walking to the gym, I just realized that this was one of the dumbest "proofs" ever, but I did observe something and jotted it down with iNotes. I thought about distance from $\mu$, the average that is always just $n$, and I think the distance is the same if we give some constraints.
Let $p_2$ be the smallest possible prime such that we can still add another prime number, $p_1$ to form an even number greater than 2. For some reason, then $p_2$ will typically be 3 or 5, sometimes 7, and should be 2 only one time for 4 = 2 + 2. Emerging Patterns (I didn't fact check each p1 and p2, but the main idea is still there)
The way we get the 2 working prime numbers is to find some number and add it to n to get the first prime number. Then subtract the same number to get the second prime number. For example: notice that n = 14, 2n = 28 = 23 + 5. We get those 2 numbers by:
- adding 9 to 14 to get 23, 23 = 14 + 9
and then
- we subtract 9 from 14 to get 5, 4 = 14 - 9
Therefore, 9 is the number that is relating 23 and 5 to 14.
It seems like if you can find a relationship between n and the number we add and subtract, then you can always find 2 prime numbers that add to the even number $2n$. From this table in this picture, p_2 is special, maybe constraining $p_2$ is very important? Maybe show that the only possible values for $p_2$ are 2,3,5 and 7?
Note: This is the answer to the question as it was asked originally. In the meantime, the OP substantially changed it.
This is not a legitimate proof. If you take it at face value, you have only proven that $0\equiv 0\pmod 2$, which you already knew all along!
What is more interesting is what led you to believe that it was legitimate. I think it may be because of the stress in the math education on calculating, rather than logic. The pattern in your proof mimics what we would normally do to solve equations such as, say, $x+\sqrt{x+5}=7$:
$$\begin{array}{rcl}x+\sqrt{x+5}&=&7\\\sqrt{x+5}&=&7-x\\x+5&=&x^2-14x+49\\0&=&x^2-15x+44\\x&=&\frac{15\pm\sqrt{15^2-4\times 44}}{2}=4\text{ or }11\end{array}$$
Oops, $x=4$ is the solution, but $x=11$ isn't! What went wrong???
The answer is: the subsequent rows in the derivation above are only consequences of the previous rows, they are not equivalent to them. In particular we made $3$rd row by squaring the second row. If two numbers are equal ($2$nd row), then their squares are equal ($3$rd row). However, if the squares of two numbers are equal, it does not follow that the numbers themselves are equal! That is why values for $x$ that are not solutions of the initial equation have crept into subsequent equations.
Back to your "proof". Each row is a consequence of the previous row, but it may not be equivalent to it. You ended up with something that is true, but to prove that the thing you've started with is true, you would need to have the subsequent rows imply the previous rows - which is not the case. In fact, at a crucial point you said "$p_1$ and $p_2$ were prime, therefore they were odd". That step is going to be the hardest (read: impossible) to reverse: surely not all odd numbers are prime.