Find all ordered pairs $(a,b)$ such that $1/a + 1/b = 3/2018$ and $a,b$ are positive integers

Question

I gave this problem my best attempt and am now trying to understand the solution for it. This is problem #1 to the 79th William Lowell Putnam Math Competition. This is the given solution by Kiran Kedlaya and Lenny Ng:

By clearing denominators and regrouping, we see that the given equation is equivalent to $(3a−2018)(3b−2018) = 2018^2$. Each of the factors is congruent to $1 \text{ (mod } 3)$. There are $6$ positive factors of $2018^2 = 2^2 · 1009^2$ that are congruent to $1\text{ (mod } 3)$: $1$, $2^2$ , $1009$, $2^2 · 1009$, $1009^2$, $2^2 · 1009^2$. These lead to the 6 possible pairs: $(a,b)$ $= (673,1358114)$, $(674,340033)$, $(1009,2018)$, $(2018,1009)$, $(340033,674)$, and $(1358114,673)$. As for negative factors, the ones that are congruent to $1\text{ (mod }3)$ are $−2$, $−2 · 1009$,$−2 · 1009^2$. However, all of these lead to pairs where $a ≤ 0$ or $b ≤ 0$.

I don't fully understand these things:

1. Each of the factors is congruent to $1\text{ (mod }3)$: I think congruence means the ability to translate into something else using some rules, so it seems like they are saying that $(3a-2018) = 1$ and $(3b-2018) = 1$. I'm also not sure why they wrote $1\text{ (mod }3)$ because 1 % 3 = 1, so why not just say $1$?
2. There are $6$ positive factors of $2018^2 = 2^2·1009^2$ that are congruent to $1\text{ (mod }3)$: $1$, $2^2$, $1009$, $2^2·1009$, $1009^2$, $2^2·1009^2$: Here it seems like they are listing all of the ways to make $2018^2$, but if they are saying that $1$ is a possible factor then wouldn't $2018^2$ be its pair and therefore must be in this list? Why wouldn't they put it into this list?
3. These lead to the 6 possible pairs:$(a,b) = (673,1358114)$, $(674,340033)$, $(1009,2018)$,$(2018,1009)$, $(340033,674)$, and $(1358114,673)$: How did they get these pairs?

Answer 1

We need to solve $$3ab=2018(a+b)$$ or $$9ab-3\cdot2018(a+b)+2018^2=2018^2$$ or $$(3a-2018)(3b-2018)=2018^2$$ or $$(3a-2018)(3b-2018)=2^21009^2.$$ Now, let $a\leq b$.

We obtain: $$3a-2018=1,$$ which gives $a=673$ and $b=1358114$ or $$3a-2018=2,$$ which is impossible because $2+2018$ is not divisible by $3$ or $$3a-2018=4,$$ which gives $a=674$ and $b=340033$ or $$3a-2018=1009,$$ which gives $a=1009$ and $b=2018$ or $$3a-2018=2\cdot1009,$$ which is impossible because $2\cdot1009+2018$ is not divisible by $3$.

Can you end it now?

Answer 2

1. Saying a number is congruent to $1 \bmod 3$ means that when you divide it by $3$ you have a remainder of $1$. The point is that $3a-2018$ is congruent to $1 \bmod 3$ because $3a$ is a multiple of $3$ and $2018$ has a remainder of $2$ when divided by $3$.
2. $2018^2$ is on the list as $2^2\cdot 1009^2$
3. They used each factorization and solved for $a$ and $b$. The factorization $1 \cdot 2018^2$ gives us $3a-2018=1,a=673, 3b-2018=2018^2,b=\frac 13(2018^2+2018)=1358114$

Answer 3

$$\frac{1}{a} + \frac{1}{b} = \frac{3}{2018} \implies 2018a +2018b -3ab = 0$$

$$\implies b = \frac{2018 a}{3 a - 2018} \land 3 a\ne 2018\quad\lor\quad a = \frac{2018 b}{3 b - 2018} \land 3 b\ne 2018 $$

Since $\quad 2018/3\not\in\mathbb{N}\quad$ we need not worry about the restrictions above. A spreadsheet or a simple program reveals the following (a,b) pairs.

$$ (-678048,672)\quad (0,0)\quad (672,-678048)\quad (673,1358114)\quad (674,340033)\\ (1009,2018)\quad (2018,1009)\quad (340033,674)\quad (1358114,673)\quad $$ These are the only pairs found for $-10^9 \le a \le 10^9$ and, based on the ascending-a and descending-b pattern in the pairs, it appears that these are the only solutions to the equation. Also, since $3$ of them are non-positive, only the $6$ you found meet the criteria.