$$ \log_2 \left(1 + \frac{1}{a}\right) + \log_2 \left(1 + \frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.} $$
Apparently, the answer is $a= 1$, $b =2$, and $c\space = 3$.
When I asked my math teacher I was told that the solution involved a bit of number theory, but didn't recieve a complete explanation. Could someone clear that up for me?
Edit: I had made a mistake in typing the question. I had left it as:
$ \log_2 \left(a + \frac{1}{a}\right) + \log_2 \left(b + \frac{1}{b}\right)+ \log_2 \left(c + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.} $
My apologies for causing confusion.
The problem, as you have written it, has no solution.
Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$ But, by the AM-GM inequality, we get $x^2+1\ge2x$, which gives $$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$