Let $a,b,c$ be nonzero integers, with $1$ as their only positive common divisor, such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$. Find the number of such triples $(a,b,c)$ with $50 \ge |a| \ge |b| \ge |c| \ge 1$.
From $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$, I know that $ab+bc+ca=0$. Thus we must have positives and negatives among $(a,b,c)$. I also know that $\gcd(a,b,c) = 1$ necessarily but I could have $\gcd(a,b)=p$, $\gcd(b,c) = q$ and $\gcd(c,a) = r$ and $p,q,r$ need not be $1$, though $\gcd (p,q,r)=1$. But how can I proceed from here? (Though the question doesn't say so explicitly I'm supposed to write a full solution/proof).