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 . And we are done.
Suppose n is congruent to 2 (mod 3). Then . And we are done.
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:
which is a multiple of 3 and thus not a prime number.
If n has a remainder of 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 ; thus, 3 consecutive odd would be of the form { }.
Case 1:
Let .
{ }; therefore, or isn't prime, since if and , then is composite.
Case 2:
Let .
{ }.
Case 3:
Let .
{ }.
I am not sure now how to tie in that it follows in case 2 and 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 ." What you really mean is "3 consecutive odd members of " 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.)