# Is the Green-Tao theorem valid for arithmetic progressions of numbers whose Möbius value $\mu(n)=-1$?

I am reading the basic concepts of the Green-Tao theorem (and also reading the previous questions at MSE about the corollaries of the theorem). According to the Wikipedia, the theorem can be stated as: "there exist arithmetic progressions of primes, with $k$ terms, where $k$ can be any natural number". There is also an extension of the result to cover polynomial progressions.

My question is very basic, but I am not sure about the answer of this:

If green-Tao theorem is true for prime numbers, then would it be true automatically for those numbers whose Möbius function value is $\mu(n)=-1$?

Something like (1):"there exist arithmetic progressions of numbers whose Möbius function value $\mu(n)=-1$, with $k$ terms, where $k$ can be any natural number"

My guess is that if all prime numbers are $\mu(p)=-1$ then the theorem is also true for those numbers whose $\mu(n)=-1$ (including primes and other numbers).

At least if the prime numbers are included (1), then I suppose that is (trivially?) true, but I am not sure if it could be said (2) that the theorem would be true for non-prime numbers whose $\mu(n)=-1$ only (something like: "it is possible to find any kind of arithmetic progression of non-primes whose $\mu(n)=-1$" ).

I guess that (1) could be correct, but (2) would be wrong. Is that right?

Thank you!

• (1) is trivially implied, as you suggest, but (2) is not trivially implied. That said, it would be very surprising if the proof of Green-Tao could not be extended to handle your case - in fact, probably even the special case (which implies your question) which covers only integers that are the products of three distinct primes. – Steven Stadnicki Jul 28 '15 at 0:15
• @StevenStadnicki thank you for the comment! it is very interesting the point you mentioned about the special case of those $n$ that are products of three distinct primes. – iadvd Jul 28 '15 at 1:00
• @StevenStadnicki as you took time to answer in the comments, please may I ask you to add this to an answer to accept and close the question? (or if you do not mind, then I will add it and close it), thank you! – iadvd Jul 29 '15 at 7:21