How many fractions make 3/2018?

There is a relevant theorem by the Czech mathematician P. Bartoš. I'm having trouble finding the original 1970 paper containing the theorem, but it's Theorem 3 in this paper. The theorem states (rephrased):

Given $m/n$ with $1\leq m\leq n$ and $\gcd(m,n)=1$ , $(x,y)$ is a positive integer solution to the equation $\frac{1}{x}+\frac{1}{y} = \frac{m}{n}$ if and only if there exist $f_1,f_2\in\mathbb{Z}^+$ such that $f_1f_2=n^2$ , $x=\frac{n+f_1}{m}$ and $y=\frac{n+f_2}{m}$ (and both of these are integers).

Applying this theorem to the case of $m=3$ and $n=2018$ , we get that $f_1f_2=2018^2=2^2\cdot1009^2$ . This leaves five possibilities (up to the order of $f_1$ and $f_2$ ):

(1) $f_1=1$ and $f_2=2^2\cdot1009^2$
(2) $f_1=2$ and $f_2=2\cdot 1009^2$
(3) $f_1=1009$ and $f_2=2^2\cdot1009$
(4) $f_1=2^2$ and $f_2=1009^2$
(5) $f_1=f_2=2\cdot 1009$

Those cases for which both $f_1+2018$ and $f_2+2018$ are divisible by $3$ will be our solutions. $2018\equiv 2\bmod 3$ , so we are looking for $f_1\equiv f_2\equiv 1\bmod 3$ . These are cases (1), (3), and (4) above. So there are $\boxed{3}$ solutions.

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We can verify that these cases yield solutions, if desired.

(1) $x=\frac{1+2018}{3} = 673$ and $y=\frac{2^2\cdot1009^2+2018}{3} = 1358114$ $\frac{1}{673}+\frac{1}{1358114} = \frac{3}{2018}$ (3) $x=\frac{1009+2018}{3} = 1009$ and $y=\frac{2^2\cdot1009+2018}{3} = 2018$ $\frac{1}{1009}+\frac{1}{2018} = \frac{3}{2018}$ (4) $x=\frac{2^2+2018}{3} = 674$ and $y=\frac{1009^2+2018}{3} = 340033$ $\frac{1}{674}+\frac{1}{340033} = \frac{3}{2018}$

Clearing the denominators and factorizing the equation gives $3ab=2018(a+b)\implies \left(a -\frac{2018}{3}\right)\left(b-\frac{2018}{3}\right)=\frac{2018^2}{9}\implies (3a-2018)(3b - 2018) = 2018^2$ It it follows that $\{3a - 2018, 3b - 2018\} \in \big\{ \{1, 2018^2\}, \{4, 1009^2\}, \{1009, 4036\} \big\},$ as the other three factorizations of $2018^2$ do not give integers for $a$ and $b$ .

This gives us the solutions: $\{a, b\} \in \big\{ \{673, 673 \cdot 2018\}, \{674, 337 \cdot 1009\}, \{1009, 2018\} \big\}.$

Note: This was the first problem on the 2018 Putnam test .

The solutions are $(2018,1009), (340033,674)$ , and $(1358114,673)$ . First, I figured that there is no solution for where $a=b$ , and that $b \lt 1346$ . Solving for $a$ , you get that $a=\frac{2018b}{3b-2018}$ . If $3b-2018$ is a factor of $2018$ , $a$ is an integer. The only times when $3b-2018$ is a factor of $2018$ , and $3b-2018$ is positive (because $a$ cannot be negative) is when $3b-2018=1$ and $3b-2018=1009$ , where $b=673,1009$ . Another way $a$ is an integer is if $3b-2018$ is a factor of $b$ or $\frac{b}{2}$ or $\frac{b}{1009}$ (as 2018's factors are 2 and 1009). The only time this happens is when $b=674$ and $3b-2018=4$ .