Prove by induction that $f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}$ is a multiple of 3 if it is an integer I'm trying to solve the following problem by induction but I'm getting stuck. 

For positive integers $a$ and $b$, define $$f(a,b) = \frac{a}{b} + \frac{b}{a} + \frac{1}{ab}.$$ If $f(a,b)$ is an integer, prove that it is a multiple of 3.

Proof by induction on a:
Base case: $a=1$
$$f(1,b) = \frac{1}{b} + \frac{b}{1} + \frac{1}{b} = \frac{2}{b} + b$$
Since $f(1,b)$ is an integer than $\frac{2}{b}$ must be an integer and $b\in\{1,2\}$ . Then $f(1,b) = 3$.
Inductive hypothesis: Assume for some integer $k \ge 1,$ $$f(k,b) = \frac{k}{b} + \frac{b}{k} + \frac{1}{kb}$$ is an integer and is a multiple of 3. 
I want to show that $f(k+1,b)$ is also a multiple of 3. Then
$$f(k+1,b) = \frac{k+1}{b} + \frac{b}{k+1} + \frac{1}{(k+1)(b)} = \cdots.$$ 
And this is where I get stuck. I know there are other ways to solve this problem but I wanted to try it by induction. Hope someone can help with this! Thanks.

Note that $$f(a,b)=\frac{a^2+b^2+1}{ab}$$ So, if $ab\mid a^2+b^2+1$, then $f(a,b)$ is an integer.
Note that $a^2\equiv0,1\pmod3$. 
If $3\mid a$, then $$a^2\equiv0\pmod3\to a^2+b^2+1\equiv1,2\pmod3$$so $3$ doesn't divide it, but $3\mid ab$. This implies that $f(a,b)$ is not an integer. The same analysis holds when $3\mid b$.
So, $f(a,b)$ can only be an integer if $3\not\mid a,b$. But, that implies that $a^2,b^2\equiv1\pmod3$, which implies that $$a^2+b^2+1\equiv3\equiv0\pmod3$$i.e., $a^2+b^2+1$ is divisible by $3$, but $3\not\mid ab$. So, $f(a,b)$, if it is an integer, divides $3$.
