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

This is meaningless.

Printable View

- Jun 3rd 2010, 10:38 AMundefined
- Jun 3rd 2010, 10:40 AMANDS!
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.

Quote:

That is why I said S is equiv and then choose the appropriate element of S.

- Jun 3rd 2010, 10:41 AMdwsmith
- Jun 3rd 2010, 10:43 AMundefined
- Jun 3rd 2010, 10:44 AMdwsmith
- Jun 3rd 2010, 10:47 AMundefined
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 $\displaystyle n \equiv n+2\ (\text{mod}\ 3)$. (Which can never be true.) - Jun 3rd 2010, 10:49 AMdwsmith
- Jun 3rd 2010, 10:52 AMdwsmith
The final question is how does this justify that there is/are no other groupings of 3 consecutive odd integers?

- Jun 3rd 2010, 10:55 AMANDS!