Jtdylktuy

This question comes from the 2018 Putnam and can be found in the following link:

https://kskedlaya.org/putnam-archive/

So far, I got the following two equations using simple algebra:

(1) $2018a = b(3a - 2018)$

(2) $2018b = a(3b - 2018)$

Then, I divide (1) by $b$ and I divide (2) by $a$ to get

(3) $frac{2018a}{b} = (3a - 2018)$

(4) $frac{2018b}{a} = (3b - 2018)$

Multiplying (3) and (4) together, I get $(2018)^2 = (3a - 2018)(3b-2018)$

So $(2018mod3)$ is congruent to $(2mod3)$ and $(2mod3)^2$ is congruent to $(1 mod3)$. In the answer provided, I'm told that each of the factors is congruent to $(1mod3)$. Do we not consider the case when both factors are congruent to $(2mod3)$ because only $2$ satisfies this requirement, and

$2$ x $2 = 4$ is definitely not equal to $2018$?

You want to look at factorizations of $2018^2$ where each term $equiv -2018 equiv 1 mod 3$. Since $2018 = 2 times 1009$ with $1009$ prime, there are not too many solutions.

Do we not consider the case when both factors are congruent to (2mod3) ...?

There is no such case since both factors in $(3a-2018)(3b-2018)$ are congruent to $1$ (mod $3$) whatever integers you take for $a$ and $b$:
$$
3n-2018=-2018=1pmod{3},quad forall nin{mathbb{Z}}.
$$

That's why the solution says so:

Take $d=gcd(a,b)$, with $a=da_2$, $b=db_2$. Then $frac 1 a + frac 1 b = frac 1 {da_2} + frac 1 {db_2}$. $a_2$ and $b_2$ are by definition coprime, so $frac {b_2+a_2}{da_2b_2}$ is a fraction in simplest form that is equal to $frac 3 {2018}$. Since $da_2b_2=2018$, and $1009$ is a prime factor of $2018$, it follows that of $d$, $a_2$, and $b_2$, one of them must be $1009$, and clearly that would be $d$ (otherwise the numerator $b_2+a_2$ would contain a $1009$, which is more than $3$). It then follows that of $a_2$ and $b_2$, one of them is $2$, and then it follows that the other is $1$.

More generally,
if
$dfrac1{a}+dfrac1{b}
=dfrac{u}{v}$,
then
$v(a+b)
=abu
$
or

$begin{array}\
0
&=abu^2-uv(a+b)\
&=abu^2-uv(a+b)+v^2-v^2\
&=(au-v)(bu-v)-v^2\
end{array}
$

so
$(au-v)(bu-v)=v^2
$.

In particular,
either
$augt v, bu gt v$
or
$ault v, bu lt v$.

We can assume that
$a ge b$.

If $a=b$
then
$(au-v)^2=v^2
$
so
$au-v = pm v$
so that
$a = frac{2v}{u}$
(which gives
$frac1{a}=frac{u}{2v}$)
or
$a=0$ ,
which not be.
For there to be a solution here,
we must have
$u | 2v$.

So we can assume
from now on
that $a > b$.

For each factorization
$v^2 = rs$
with $r le s$,
for a solution
we must have either
$au-v=r, bu-v = s
$,
$au-v=s, bu-v = r
$,
$au-v=-r, bu-v = -s
$,
or
$au-v=-s, bu-v = -r$.

Since
$a > b$ and $r le s$,
the first and third cases can not hold,
so either
$a = frac{v+s}{u},
b = frac{v+r}{u}
$
or
$a = frac{v-r}{u},
b = frac{v-s}{u}
$.

For these to be integers,
we must have either
$u | (v+s),
u | (v+r)$
or
$u | (v-r),
u | (v-s)$.
In either case,
we must have
$u | (s-r)$.

By looking at
$v bmod u$
we can quickly decide
which $r$ and $s$
will work.