Say $f(x)=\dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number. So I reframe $f(x)$ to
$\dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$\dfrac{3x-1}{4}=k_1$ or $\dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$ How do I frame the equation?
I tried putting
$\dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= \dfrac{3(\frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.
Continuing what you have written,
$f^2(x)=\frac{9x+5}{4}=2k+1$
$k=\frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=\frac{9(7+y)+1}{8}$$ $$k=8+\frac{9y}{8}$$ $$\implies 8|y $$.
It gives $x=7,15,23,31...$