Finding the x's for which f(x) is a whole number I'm trying to improve my math skills and I found this exercise in my book. 
We have the following function: $$f: \mathbb R\to \mathbb R$$  $$f(x) = \frac {4x+1}3$$ I've been asked to find out one value for $x$ such that $f(x)$ is a whole number. I found some solutions and afterwards I determined that for all $$x_n=\frac {3^{2n}-1}4$$  where n is a natural number different from 0 , $f(x_n)$ is a whole number. I'm not 100% sure this is correct but it seems to work for all positive even powers of 3. 
I've been trying to prove that this formula is corect, but I didn't succeed. Could anyone point me out on how to prove this kind of problems, and what to watch out for? Thanks. 
 A: To see that your expression always works, note that $$3^{2n}=(3^2)^n=9^n=(1+8)^n=1+n\times 8 + \binom n2\times 8^2 +\cdots + 8^n\implies$$$$3^{2n}-1= n\times 8 + \binom n2\times 8^2 +\cdots + 8^n$$ which is manifestly divisible by $4$, and even divisible by $8$.
More broadly, though any integer of the form $x=3k+2$ (for integer $k$) will work.
A: $$ (4x+1)/3 = k $$
$$ (4x+1)=3k$$
$$ 4x=3k-1 $$
$$ x=(3k-1)/4$$
Thus for all integers  $ k \ge 0 $ if  you let $x=(3k-1)/4 $, you will get $f(x)=k$ which is a whole number. 
A: For $\frac {4x + 1}3= m \in \mathbb N$
Then $4x + 1 = 3m$
$4x = 3m -1$
$x = \frac {3m -1}4$.  Now $x$ doesn't have to be a whole number so for $m = 0,1,2,3,4,5.....$ we get $x = \frac {-1}{4}, \frac {1}{2}, 2, \frac {11}4, \frac 72....$  and $f(x) = 0,1,2,3,4,5..... $ etc.
Now if $m = 3^{2n-1}$ then $x = \frac {3*3^{2n-1} -1}4 = \frac {3^{2n} -1}4$ is certainly acceptable but very very specific.
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In other words:
$\frac{4(\frac {3^{2n} - 1}4) + 1}{3}=\frac {(3^{2n}- 1) + 1}3=\frac {3^{2n}}3 = 3^{2n-1}$ is certainly a whole number.
But so is $\frac {4(\frac {3m-1}4) + 1}3 = \frac {(3m-1)+1}3 = \frac {3m}3 = m$.
A: You have found some of the solutions... you have not found all of the solutions.
If $f(n)$ is a solution
$f(n +3) = \frac {4(n+3) + 1}{3} = \frac {4n + 1}{3} + 4$
And so once you find one, then you know that adding $3$ will give another.
So will subtracting $3.$
Another way to think about it...
$f(x) = \frac {4x + 1}{3}\\
f(x) = \frac {3x+x + 1}{3}\\
f(x) = x + \frac {x + 1}{3}$
And so then you need to find those $x$ such that the last term is an integer.
