If you stick strictly with a direct proof (denoting two consecutive integers by $n$ and $n+1$, summing them to get $2n + 1$, therefore odd), you'll be fine.

For one thing, your assumed contradiction, the negation of "the sum of two consecutive numbers is always odd" is not correctly stated; its negation needs to be "it is not the case that that the sum of two consecutive numbers is always odd", which means, "there *exists* two consecutive integers whose sum is even."

Proof by contradiction here turns out to be much more work than
simply using a direct proof.

Your teacher may have chosen to represent two cases, in the event some students designated the two consecutive integers as $(n - 1)$ and $n$, while others, like you, denoted these consecutive integers as $n$ and $(n+1).$

As you note, the proof proceeds the same, in either case, but given that your teacher was providing an answer key, he or she may have simply tried to cover all the bases: all the approaches students may have used.

But to answer your question, aside from this pedagogical concern your teacher may have had, the proof does not require a proof by cases. So I do not believe your teacher expected you or any other student to provide both cases. Doing so does not add any more information, is rather redundant, etc, save for the pedagogical concerns your teacher may have had.

Is it a tautology? Yes

Why? Just draw the truth table and see that the truth value of the main implication is always 'true'. Here, I say '0' is 'false' and '1' is true:
\begin{array}{|c|c|c|c|c|}
\hline
p & q & p\to q & p\land(p\to q) & (p\land(p\to q))\to q \\ \hline
0 & 0& 1&0&1\\ \hline
0 & 1& 1&0&1\\ \hline
1 & 0& 0&0&1\\ \hline
1 & 1& 1& 1& 1\\ \hline
\end{array}

Edit: you can also say
\begin{align}p \land (p \to q) &\equiv p \land (\lnot p \lor q)\\ &\equiv (p \land \lnot p) \lor (p \land q )\\ &\equiv \text {false} \lor (p \land q)\\&\equiv p \land q, \end{align}

that implies $q $.

Edit 2: this inference method is called *modus ponens*, and it is the simplest one.

For instance, say $p = $ it rains, $g = $ I get wet.

So, if we know that the implication *if it rains, then I get wet* is true (that is, $p\to q $) and *It rains* ($p $), what can we infere? It is obvious that *I get wet* ($q $).

Note that, even though $p \land (p\to q) $ is not verified, as $p\land (p\to q) \to q$, the main implication is verified.

## Best Answer

If you want to do a proof by cases, the first step is to identify a complete list of possible cases (in principle, they need not be mutually exclusive, but in practice they usually are).

Once you have your completely list of possible cases, then you examine each case in turn. If in all cases you get the desired conclusion, then the conclusion holds in all cases.

So, here you want to show that if $3$ does not divide $n$, then $3$ divides $n^2-1$, and you want to do it "by cases". So, what is the complete list of "cases" that covers all possibilities? "Cases" for what or whom?

Since the only thing that can change from one instance of this problem to the other is $n$, then the cases will refer to $n$. There are many ways to break up all possible $n$ into cases. The simplest here, because you are dealing with divisibility by $3$, is to consider the possible remainders when you divide $n$ by $3$.

So what are all the possible situations when you divide $n$ by $3$? You can get a remainder of $0$, a remainder of $1$, or a remainder of $2$. So we go through each case in turn.

So, suppose that $3$ does not divide $n$; we want to conclude that $3$ divides $n^2-1$.

CASE 1: The remainder is $0$. This case is impossible, because in this case $3$ divides $n$, and we are assuming that $3$ does not divide $n$. So we can discard this case.

CASE 2: The remainder is $1$. Then $n$ can be written as $n=3k+1$; square both sides, subtract $1$, see what happens.

CASE 3: The remainder is $2$. Then $n$ can be written as $n=3k+2$;

`<continue here>`

.Since, in all cases we get the conclusion we want, the conclusion holds for all $n, as desired.