[Number Theory]
Problem: For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.

For example, the value of $f(1,2)$ is 3 .
Determine all positive integers $a$ for which $f(a, a)$ is an integer.

Solution: By definition, $f(a, a)=\frac{a}{a}+\frac{a}{a}+\frac{1}{a^{2}}=2+\frac{1}{a^{2}}$.

For $2+\frac{1}{a^{2}}$ to be an integer, it must be the case that $\frac{1}{a^{2}}$ is an integer.

For $\frac{1}{a^{2}}$ to be an integer and since $a^{2}$ is an integer, $a^{2}$ needs to be a divisor of 1 .

Since $a^{2}$ is positive, then $a^{2}=1$.

Since $a$ is a positive integer, then $a=1$.

Thus, the only positive integer $a$ for which $f(a, a)$ is an integer is $a=1$.

Final answer: 1