I would like to know whether I have correctly proved the following result.
My book gave a different proof. My concern is primarily with correctness and notation. Also, we are given: if $3 \nmid n$ then $n=3k+1$ or $n=3k+2$ and so $3\mid(n^2-1)$. $k \in \mathbb{Z}$
$$\text{Let} \space n \in \mathbb{Z}, \text{ Prove that } 3\mid(2n^2+1) \text{ if and only if } 3 \nmid n.$$
$$\text{Proof: We first prove the if direction of the biconditional, } 3\nmid n\implies 3\mid(2n^2+1) \\ \text{Hence, we have the following two cases. } $$
$$\text{1: n=3k+1} \\ \text{2: n=3k+2}\\ \text{Case 1: }2n^2+1=2(3k+1)^2+1= 3(6k^2+4k+1)\\ \text{Since } 6k^2+4k+1 \text{ is an integer } \\ \therefore 3\mid(2n^2+1) \\ \text{Case 2: Similar, therefore omitted.}$$
$$\text{Finally, for the if only direction we use the contrapositive, } 3\mid n\implies 3\nmid(2n^2+1)$$ $$\text{We know $3\mid n$. Thus, } n=3k, k\in \mathbb{Z} \\ \text{Then, } 2n^2+1=3(6k^2)+1\\ 6k^2 \text{ is an integer } \\ \therefore 3\nmid(2n^2+1)$$
$$\text{Now we conclude that: }3\mid(2n^2+1) \text{ if and only if } 3 \nmid n \text{ } \blacksquare$$
I might use too many words but I try to be as clear as possible since I am still learning how to do proofs correctly.
Best Answer
Your proof looks good. If this is for an introduction to proofs, it is always helpful to be explicit when you are starting out even if it is redundant. You never know if there is redundancy until you try. But your proof is correct.