High school student here...
Ok, so I've been working on how to create proofs this summer and I think I've finally completed one correctly. I would like feedback on whether I've done this correctly and if there is a better way to word certain parts of the proof. One of my biggest hurdles so far has been terminology.
While playing around with odd numbers one day I came across a pattern: $1+3 = 4, 5+7 = 12, 9+11 =20...$
I realized that for every pairing of consecutive odd numbers, the sum is a multiple of 4. I'm aware that this is most likely far from new information, but I thought it would be fun to try and prove it. After many hours of learning basic logic and proof "structure", this is what I came up with:
"Conjecture":
Assume that $n$ is an odd integer denoted, $n=2k+1$ for some integer $k$. For any pair of consecutive odd numbers $n$, $(n+2)$, the sum is a multiple of $4$.
Proof:
We can assume that $n$ is an odd integer $2k+1$ as stated by the conjecture. We are trying to prove that: $$n+(n+2)=4n$$
We can rewrite our equation as: $$(2k+1)+(2k+3)=4n$$
which then can be simplified to: $$4k+4=4n$$
Now using algebraic manipulation, we find that $$n=(k+1)$$
We can plug this information into our original equation: $$4k+4 = 4(k+1)$$
Simplified, we have the equality: $$4k+4=4k+4$$
Therefore, the sum of two consecutive odd numbers $n+(n+2)$ is always equivalent to a multiple of $4$.
This is definitely the right idea but there's a minor problem.
First, you do not want to prove $$ n + (n+2) = 4n.$$ Notice the $n$ on both sides. This would mean, taking $n=3,$ $$ 3 + (3+2) = 4\cdot 3 = 12,$$ which is is wrong.
You really want to prove that $$ n + (n+2) = 4m$$ for some integer $m$.
Second - and this is partially about style as well as substance - you don't want to start your proof with the equation that you want to prove. This can be confusing even if it amounts to a correct proof, and can also invite logical errors like circular reasoning and affirming the consequent.
Instead, try a chain of equalities. I would rewrite your proof as $$n + (n+2) = 2k+1 + (2k+3) \\= 4k + 4 \\= 4(k+1). $$ This shows that $n + (n+2)$ is four times $k+1,$ and thus a multiple of four.