Prove that 3, 5, 7 are the only three consecutive odd integers that are primes.

How to start?

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- Jun 3rd 2010, 10:12 AMdwsmith[SOLVED] 3, 5, 7 primes
Prove that 3, 5, 7 are the only three consecutive odd integers that are primes.

How to start? - Jun 3rd 2010, 10:20 AMundefined
- Jun 3rd 2010, 10:22 AMdwsmith
- Jun 3rd 2010, 10:27 AMundefined
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. - Jun 3rd 2010, 10:41 AMdwsmith
- Jun 3rd 2010, 10:45 AMundefined
- Jun 3rd 2010, 10:52 AMdwsmith
- Jun 3rd 2010, 10:57 AMundefined
- Jun 3rd 2010, 11:05 AMANDS!
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. - Jun 3rd 2010, 11:16 AMdwsmith
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? - Jun 3rd 2010, 11:26 AMANDS!Quote:

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. . .}. - Jun 3rd 2010, 11:30 AMdwsmith
- Jun 3rd 2010, 11:31 AMundefined
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.) - Jun 3rd 2010, 11:34 AMundefined
- Jun 3rd 2010, 11:35 AMdwsmith