Integer equation with one parameter I need to find positive integers $f = f(n)$ and $g = g(n)$ both dependent on $n \in \mathbb N$ so that
$$ \frac1g + \frac1f = \frac{3}{3n-2} $$
for all $n$ (Or at least all $n>N$ where $N$ is a arbitrary bound.)
I have no clue how to even begin to tackle this problem.  The only thing up to now is $\displaystyle\frac1g + \frac1f = \frac{f+g}{fg}$ which is pretty trivial.
 A: If $n$ is even ($n=2q$), then $f(2q)=3q-1$ and $g(2q)=2(3q-1)$ are working.
If $n$ is odd ($n=2q+1$) and the problem is equivalent to $\frac{1}{f}+\frac{1}{g}=\frac{3}{6q+1}$. 
Wolfram was not able to find a solution for all the random $q$ I tried so I am not sure there is a solution, even for large $n$. 
But maybe @user159870 method will be able to prove it.
A: If you mean $n$ instead of $k$:
Edit:
$$\frac{1}{f}+\frac{1}{g}=\frac{3}{3n-2} \Rightarrow (3n-2)(g+f)=3fg $$
$$3 \mid (3n-2)(g+f) \Rightarrow 3 \mid 3n-2 \text{ or } 3 \mid g+f$$
$$3 \mid 3n \text{ if } \Rightarrow 3 \mid 3n-2 \text{ then } 3 \mid 3n-(3n-2) \Rightarrow 3 \mid 2 $$
So $$3 \mid f+g \Rightarrow f+g=3m \Rightarrow f=3m-g$$
Replace this in $(3n-2)(g+f)=3fg$
$$(3n-2)(g+3m-g)=3(3m-g)g \Rightarrow (3n-2)3m=9mg-3g^2 \Rightarrow \\ 3g^2-9mg+(3n-2)3m=0$$
SOlve for $g$ using the discriminant.
Then replace this at $f=3m-g$ to find $f$.
A: if $n = 3$ you get a target sum of $3/7$ and clearly there is no solution of the form $1/f + 1/g$ where $f$ and $g$ are positive integers. In general, any time $p = 3n - 2$ is prime there shouldn't be any solutions to get a target sum of $3/p$.
A: For the equation;  $$\frac{1}{a}+\frac{1}{b}=\frac{3}{3q-2}$$ 
Degradable number of multipliers.  $$(k-3t)(k+3t)=4(3q-2)^2$$ 
Find.  $k,t$ Substitute in the formula and solutions are. 
$$a=q+\frac{\pm{k}-4-3t}{6}$$
$$b=q+\frac{\pm{k}-4+3t}{6}$$
What's difficult could that be?
Let  $q=4$ Then:  $3q-2=10$
$$(k-3t)(k+3t)=4*10^2=10*40$$
$$k=3t+10$$
$$6t=40-10$$
$$t=5$$
$$k=25$$
$$a=4+\frac{25-4-15}{6}=5$$
$$b=4+\frac{25-4+15}{6}=10$$
$$\frac{1}{5}+\frac{1}{10}=\frac{3}{10}$$
