# Thread: [SOLVED] 3, 5, 7 primes

1. Originally Posted by dwsmith
I don't like the confusion of n and then having n+2 and n+4. That is why I set S is equiv and then choose the appropriate element of S.
You have a set congruent to an integer modulo another integer. This makes no sense. For example, let A = {1,2,3,4,5,8}.

Solve for x: $A \equiv x\ (\text{mod}\ 3)$.

This is meaningless.

2. Originally Posted by dwsmith
If $n+2\equiv 1 \ (\text{mod} \ 3)$, then $n\neq 1$ since $1+2=3\not\equiv 1 \ (\text{mod} \ 3)$
My apologies. I do not mean that N+2 is congruent to 1(mod3); I mean that if you suppose that N is congruent to 1(mod3) (and thus is of the form 3k+1), then N+2 will be congruent to 0(mod3), which would make your second odd integer not prime. The same reasoning follows if N is congruent to 2(mod3), N+4 will fail to be prime.

These are the three cases Undefined is speaking of.

That is why I said S is equiv and then choose the appropriate element of S.
You do not have to do all that though. What Undefined (I believe) is trying to convey, is that your starting odd integer has to belong to one of three classes: 0, 1 or 2(mod3). It can not belong to 0, as that would make it non-prime. If it belongs to 1(mod3), then one of your other integers will fail (the second one). If it belongs to 2(mod3) then the other integer will fail (the third one). Your then done, as you've shown that all cases result in a list that includes at least one non-prime.

3. Originally Posted by undefined
You have a set congruent to an integer modulo another integer. This makes no sense. For example, let A = {1,2,3,4,5,8}.

Solve for x: $A \equiv x\ (\text{mod}\ 3)$.

This is meaningless.
But why you said $n\equiv \dots$ and then picked {n,n+2, or n+4} to take the place of n that to me makes no sense because n=n+2 for the substitution to work.

4. Originally Posted by dwsmith
But why you said $n\equiv \dots$ and then picked {n,n+2, or n+4} to take the place of n that to me makes no sense because n=n+2 for the substitution to work.
I did not substitute. I was stating an implication ( $\Longrightarrow$).

IF n is congruent to 1 (mod 3), THEN n + 2 is congruent to 0 (mod 3).

5. Originally Posted by undefined
I did not substitute. I was stating an implication ( $\Longrightarrow$).

IF n is congruent to 1 (mod 3), THEN n + 2 is congruent to 0 (mod 3).
Look at post 7 and 8 when I asked about the substitution.

6. Originally Posted by dwsmith
Look at post 7 and 8 when I asked about the substitution.
Oh, I was reading "sub in" loosely as in,

first we had n to the left of the congruence sign

then we had n + 2 to the left of the congruence sign

so n + 2 was "subbed into" the place holder that is on the left of the congruence sign.

I certainly did not mean to say that $n \equiv n+2\ (\text{mod}\ 3)$. (Which can never be true.)

7. Originally Posted by undefined
Oh, I was reading "sub in" loosely as in,

first we had n to the left of the congruence sign

then we had n + 2 to the left of the congruence sign

so n + 2 was "subbed into" the place holder that is on the left of the congruence sign.

I certainly did not mean to say that $n \equiv n+2\ (\text{mod}\ 3)$. (Which can never be true.)
That is why I decided to change n equiv to set with elements to avoid such issues.

Saying x is equiv such that x is some integer and then picking n, n+2, n+4 may be a better method.

8. The final question is how does this justify that there is/are no other groupings of 3 consecutive odd integers?

9. By this bit here, with n as an arbitrary starting integer:

Originally Posted by undefined
It is easiest to make the three cases:

Case 1: $n \equiv 0\ (\text{mod}\ 3)$

Case 2: $n \equiv 1\ (\text{mod}\ 3)$

Case 3: $n \equiv 2\ (\text{mod}\ 3)$
All cases lead to a contradiction of the assumption that there is a set of consecutive odd prime triples that is not "3, 5, 7".

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