12. Let $a, b, c$ be positive integers such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ and $\operatorname{gcd}(a, b, c)=1$. Suppose $a+b \leq 2011$. Determine the largest possible value of $a+b$.
12. Answer. 1936

Solution. We first show that $a+b$ must be a perfect square. The equation $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ is equivalent to $\frac{a-c}{c}=\frac{c}{b-c}$. Write $\frac{a-c}{c}=\frac{c}{b-c}=\frac{p}{q}$, where $\operatorname{gcd}(p, q)=1$. From $\frac{a-c}{c}=\frac{p}{q}$, we have $\frac{a}{p+q}=\frac{c}{q}$. Since $\operatorname{god}(p, q)=1$, we must have $q$ divides $c$. Similarly from $\frac{b-c}{c}=\frac{q}{p}$, we have $\frac{b}{p+q}=\frac{q}{p}$. Since $\operatorname{gcd}(p, q)=1$, we must have $p$ divides $c$. Thus $\operatorname{god}(p, q)=1$ implies $p q$ divides $c$. Therefore $\frac{a}{p(p+q)}=\frac{b}{q(p+q)}=\frac{c}{p q}$ is an integer $r$. Then $r$ divides $a, b$ and $c$, so that $r=1$ since $\operatorname{gcd}(a, b, c)=1$. Consequently, $a+b=p(p+q)+q(p+q)=(p+q)^{2}$.
Next the largest square less than or equal to 2011 is $44^{2}=1936$. As $1936=1892+44$, and $\frac{1}{1 \mathrm{~s}}+\frac{1}{44}=\frac{1}{43}$, where $\operatorname{god}(1892,44,43)=1$, we have $a=1892, b=44$ and $c=43$ give the largest value of $a+b$. These values of $a, b, c$ can obtained from the identity $\frac{1}{m^{2}-m}+\frac{1}{m}=\frac{1}{m-1}$.