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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.

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    $\begingroup$ $f\circ f(x)=\frac {9x+5}4$. Using that, the method you used first works. $\endgroup$ – lulu Dec 8 '18 at 10:52
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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 $x=7,15,23,31...$

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