Q2020/14) Solve in natural numbers $\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}$

we get
$2018b + 2018 a = 3ab$
or $3ab - 2018 a - 2018b = 0$
As the coefficient of a,b are same so we get product of 2 monomials one of a one of b and coefficients being same.
Multiplying by 3 we get
$9ab - 3 * 2018 a - 3 * 2018b = 0$
or $9ab - 3 * 2018 a - 3 * 2018b = 2018 ^2 $
or $(3a-2018)(3b-2018) = 2018^2= 2^2 * 1009^2$
LHS each term is 1 mod 3 so we should make the RHS so product of 2 numbers each of which is 1 mod 3.


if(a,b) is a solution then (b,a) is also a solution so let us assume that $a>=b$

so we get following sets

$3a-2018= 2018^2, 3b-2018=1$  giving $(a= 1358114,b= 673)$
$3a-2018= 1009^2 , 3b=2018=4$ giving $(a= 340033, b= 674)$
$3a-2018= 1009, 3b-2018=1009 * 4$ giving $(a= 1009,b= 2018)$

and permutation of the same.