Start with $1\in Q$
Applying the other conditions to $\{ 1\}$, we see that $3\in Q$.
So $\{1,3\}\subseteq Q$.
Applying the conditions to each new element of the above, we see $ 5 $ and $7$ are in $Q$.
So $ \{1,3,5,7\}\subseteq Q. $
Applying the conditions to each new element of the above, we see $ 9 $, $11$, $13$, and $15$ are in $Q$.
So $ \{1,3,5,7, 9, 11, 13, 15\}\subseteq Q. $
The next step will give you
$ \{1,3,5,7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31\}\subseteq Q. $
And you can continue in this manner to show that $Q$ contains all odd integers. Since the conditions always give odd integers, $Q$ is in fact the set of odd integers. (That is, if $Q$ is determined by the conditions. As stated, there is the possibility that other even numbers are in it. But the conditions won't give new even numbers).