exist infinitely many positive integers $n$ such $\omega (n)+\omega (n+1)\equiv 0\pmod 3$ For integer $n>1$, $\omega (n)$ denotes number of distinct prime factors of $n$, and $\omega (1)=0$. Prove that:there exist infinitely many positive integers $n$ satisfying 
$$\omega (n)+\omega (n+1)\equiv 0\pmod 3$$
Positive integers $n$ satisfying $\omega(n)<\omega(n+1)<\omega(n+2)$

With $\Omega$ instead of $\omega$ I have a solution. 
For any $n$ then $a = \Omega(2n) \bmod 3,b=\Omega(2n+1)\bmod 3,c=\Omega(2n+2)\bmod 3$, $\Omega((2n+1)^2) = 2b\bmod 3, \Omega((2n+1)^2-1) = a+c\bmod 3,\Omega(n)+\Omega(n+1) = a+c-2 \bmod 3$

Assume we chose $n$ such that $a=2\bmod 3$

The claim $\Omega(l)+\Omega(l+1) = 0 \bmod 3$ for some $l \in \{ 2n,2n+1, (2n+1)^2-1,n\}$ would follow from 

$2+b=0\bmod 3$ or $b+c=0\bmod 3$ or $2+c+2b=0\bmod 3$ or $2+c-2=0\bmod 3$.

The only fail is when $c=2\bmod 3$.
But if $c=2\bmod 3$ we can redo it replacing $n$ by $n+1$, finding that for the claim to fail we need $\Omega(2(n+m)) = 2\bmod 3$ for every $m\ge 0$, a contradiction.
