Prove that 3, 5, 7 are the only three consecutive odd integers that are primes.
How to start?
Sure.
Three consecutive odd integers can be written n, n + 2, n + 4. We know n is of the form n = 2k + 1, but this is not important for our proof.
n is either congruent to 0, 1, or 2 (mod 3). This makes 3 cases.
Suppose n is congruent to 0 (mod 3). Then it is a multiple of 3. If it is not 3, then it is not prime, and we are done.
Suppose n is congruent to 1 (mod 3). Then $\displaystyle n + 2 \equiv 1 + 2 \equiv 3 \equiv 0\ (\text{mod}\ 3)$. And we are done.
Suppose n is congruent to 2 (mod 3). Then $\displaystyle n + 4 \equiv 2 + 4 \equiv 6 \equiv 0\ (\text{mod}\ 3)$. And we are done.
Looking at this first case:
$\displaystyle n\in\mathbb{Z}^+, \ n=2k+1$ and three consecutive odd integers would be of the form $\displaystyle n, n+2, n+4$
Suppose $\displaystyle n\equiv 0 \ (\text{mod} \ 3)$
$\displaystyle n=${$\displaystyle 3,6,9,12,...$}
This is understandable but then don't I have to then show 3 would be the only odd prime out of the set that satisfies this congruence?
This was a problem we did, the reasoning is the same but does (explicitly) use modulus and congruence to prove:
As stated, three consecutive integers are of the form: n, n+2, n+4. The first integer is divisible by three with a remainder of 0, 1, or 2. 0 is obviously out.
If n has a remainder of 1:
$\displaystyle n = 3k+1 \Rightarrow n+2 = (3k+1)+2 \Rightarrow n+2 = 3(k+1);$ which is a multiple of 3 and thus not a prime number.
If n has a remainder of 2:
$\displaystyle n = 3k+2 \Rightarrow n+4 = (3k+2)+4 \Rightarrow n+4 = 3(k+2);$ which is a multiple of 3 and thus not a prime number.
Obviously not as succinct and to the point as Undefined's proof, but if you are in the same spot I was in when this question came up, this was what our Professor expected us to know.
Let $\displaystyle n\in\mathbb{Z}^+ \ \text{such that} \ n=2m+1$; thus, 3 consecutive odd $\displaystyle \mathbb{Z}$ would be of the form $\displaystyle S=${$\displaystyle n, n+2, n+4$}.
Case 1: $\displaystyle S\equiv\ 0\ (\text{mod}\ 3)$
Let $\displaystyle n\equiv\ 0\ (\text{mod}\ 3)$.
$\displaystyle n=${$\displaystyle 3,6,9,...$}; therefore, $\displaystyle n=3$ or $\displaystyle n$ isn't prime, since if $\displaystyle n=3m$ and $\displaystyle m>1$, then $\displaystyle n$ is composite.
Case 2: $\displaystyle S\equiv\ 1\ (\text{mod}\ 3)$
Let $\displaystyle n+2\equiv\ 1\ (\text{mod}\ 3)$.
$\displaystyle n=${$\displaystyle 2,5,8,...$}.
Case 3: $\displaystyle S\equiv\ 2\ (\text{mod}\ 3)$
Let $\displaystyle n+4\equiv\ 2\ (\text{mod}\ 3)$.
$\displaystyle n=${$\displaystyle 1,4,7,...$}.
I am not sure now how to tie in that it follows $\displaystyle n=5$ in case 2 and $\displaystyle n=7$ in case 3.
Obviously since we are looking for 3 consecutive odd primes it must be those.
Also, how would I then show that these are the only 3 consecutive?
By exhaustion of those two cases. If N is congruent to 1(mod 3), then N+2 is congruent to 0(mod3); and if N is congruent to 2(mod3), then N+4 is congruent to 0(mod3).Also, how would I then show that these are the only 3 consecutive?
Also 1(mod3) should be {1, 4, 7. . .} and 2(mod3) should be {2, 5, 8. . .}.
Your proof has some problems, such as confusion between sets and integers. I would avoid setting an integer equal to a set unless you're doing analysis and defining the integers, etc. I would also avoid the shorthand "3 consecutive odd $\displaystyle \mathbb{Z}$." What you really mean is "3 consecutive odd members of $\displaystyle \mathbb{Z}$" but it is easier to just write "3 consecutive odd integers."
It might help you to let n > 3. (The case n = 1 is easily dealt with separately.)