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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.

Cro-Magnon
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    If you are going to be so verbose for the sake of clarity then please say "than there exists an integer $k$ so that $n = 3k + 1$". Don't just introduce $k$ without explaination. – fleablood Jul 02 '18 at 15:38
  • @fleablood You're right. Thanks for the tip. – Cro-Magnon Jul 02 '18 at 15:41
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    Can combine the two cases Case 2: $n = 3k -1$ so $n = 3k \pm 1$ so $2(3k\pm 1) - 1 = 2(9k^2 \pm 6k + 1) -1 = 3(6k^2 \pm 4k) +1$. Can do all three cases $n = 3k + b$ for $b = 0,1,-1$ so $2n^2 - 1 = 18k^2 + 12kb + b^2 - 1 =3(6k^2 + 4kb) + b^2 -1$. And $b^2 -1 = 1$ if $b = 1,-1$ and $b^2 -1 = -1$ if $b = 0$. – fleablood Jul 02 '18 at 15:44

2 Answers2

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This is correct. You're right to use lots of words when starting to learn this kind of writing. That said, good for you for omitting case 2.

Ethan Bolker
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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.

Andrew Shedlock
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