A positive integer is *multiplicatively even (odd)* if, when decomposed into primes, the sum of the exponents is even (odd).
A subset of the integers is *thick* if it contains arbitrarily long intervals $\{n,n+1,\cdots,n+m\}$.


Is the set of multiplicatively even numbers thick?


It is easy to see that the set of multiplicatively even numbers is thick if and only if its complement (i.e. the set of multiplicatively odd numbers) is.



Is the set of multiplicatively even numbers thick?
You are asking about sign patterns of the [Liouville function](http://en.wikipedia.org/wiki/Liouville_function). While conjecturally all finite sign patterns appear infinitely often, this has been surprisingly hard to establish. One of the best results thus far is [by Hildebrand](http://www.ams.org/mathscinet-getitem?mr=874689), who showed that every sign pattern of length three occurs infinitely often, in particular the set of "multiplicatively even" or "multiplicatively odd" numbers contain infinitely many intervals of the form $\{n,n+1,n+2\}$. The length four analogue of this remains open.


On the other hand, [Szemeredi's theorem](http://en.wikipedia.org/wiki/Szemer%C3%A9di%27s_theorem) (and the prime number theorem) implies that both the set of multiplicatively even numbers and multiplicatively odd numbers contain arbitrarily long arithmetic progressions. (In fact, by using some rather complicated [results of myself, Ben Green](http://www.ams.org/mathscinet-getitem?mr=2877066), [and Tamar Ziegler](http://www.ams.org/mathscinet-getitem?mr=2950773), one can in fact show that all sign patterns occur for the Liouville function in arithmetic progressions, extending a [previous result of Buttkewitz and Elsholtz](http://www.ams.org/mathscinet-getitem?mr=2855791), though this does not directly help with the current question.)


By the way, I don't see the "easy" argument you claim that the set of multiplicatively even numbers is thick if and only if the set of multiplicatively odd numbers is. Could you elaborate? (The only thing I can think of is that you are working with all integers and considering -1 to be a prime, but it appears that you are explicitly restricting attention to positive integers in your question.) Never mind, it is obvious - every interval of length $2k$ contains the double of an interval of length $k$.


