# Arriving at odd possible solutions for functions

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.

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

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 $$x=7,15,23,31...$$