Unique Solution to a System of Equivalences 
Show that $(2, 3, 7)$ is the only solution set to the following system of equivalences:
  \begin{align}
ab = -1 & \mod c \\
ac = -1 & \mod b \\
bc = -1 & \mod a \\
\end{align}

I already attempted systems of equations like $ab + cm = -1$, $ac + bn = -1 \ldots$ but each time I was just running in circles.  So far, I know that $a, b, c$ are pairwise relatively prime.  Moreover, I must have at least two odd numbers.  Otherwise, the two even numbers are not relatively prime.  What other facts can I use in this question?

I suppose you don't count solutions like $(1,1,1)$ although they are technically solutions to the system. So for now assume w.l.o.g. $1<a<b<c$.
The equations are equivalent to $c \mid ab+1,b \mid ac+1,a \mid bc+1$.
Since (as you stated) $a,b,c$ clearly must be pairwise relatively prime, we conclude that
$$abc \mid (ab+1)(ac+1)(bc+1)=a^2b^2c^2+(abc)(a+b+c)+ab+ac+bc+1$$ and hence $abc \mid ab+ac+bc+1$.
Now, let $$\frac{ab+ac+bc+1}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=n \in \mathbb{N}$$
Then $n <\frac{4}{a} \le \frac{4}{2}=2$ and hence $n=1$ i.e. $ab+ac+bc+1=abc$.
Also $a<4$ and hence $a=2$ or $a=3$.
If $a=3$ you have $b \ge 4$ and $c \ge 5$ and hence $$1=n \le \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{60}<1$$ Absurd!
Hence we must have $a=2$.
Then it follows that $bc+1+2b+2c=2bc$ i.e. $bc-2b-2c-1=0$ or $(b-2)(c-2)=5$.
Hence $b-2=1$ and $c-2=5$ which gives exactly the solution $(a,b,c)=(2,3,7)$ you stated above...
