Solution 1: Let d=gcd(a,b), a=da′, b=db′. The given equation reduces to da′1+db′1=20211 which is equivalent to
2021(a′+b′)=da′b′.(4)
As a′ and b′ are relatively prime to each other, they both are relatively prime to a′+b′, implying that a′b′ and a′+b′ are relatively prime. By (4), a′b′∣2021(a′+b′), implying a′b′∣2021. As 2021=43⋅47 where the factors are prime, the number 2021 has exactly four positive factors 1, 43, 47 and 2021. Taking into account that a≥b holds if and only if a′≥b′, consider all cases:
* If a′=2021 and b′=1 then (4) implies d=2022. Consequently, (a,b)=(2021⋅2022,2022).
* If a′=47 and b′=43 then (4) implies d=90. Consequently, (a,b)=(47⋅90,43⋅90).
* If a′=47 and b′=1 then (4) implies d=43⋅48. Consequently, (a,b)=(2021⋅48,43⋅48).
* If a′=43 and b′=1 then (4) implies d=47⋅44. Consequently, (a,b)=(2021⋅44,47⋅44).
* If a′=1 and b′=1 then (4) implies d=2021⋅2. Thus (a,b)=(2021⋅2,2021⋅2).
Solution 2: The given equation is equivalent to aba+b=20211 which reduces to ab−2021a−2021b=0. After adding 20212 to both sides and factorizing in the l.h.s, we get
(a−2021)(b−2021)=20212.(5)
As both a and b are positive, the factors in the l.h.s. of (5) are greater than −2021. Thus if both factors were negative then the absolute value of their product would be less than 20212 and (5) could not hold. Hence both factors are positive. Since a≥b, we must have a−2021≥b−2021. As 2021=43⋅47 with factors being prime, we obtain variants (a−2021,b−2021)=(20212,1), (a−2021,b−2021)=(2021⋅47,43), (a−2021,b−2021)=(2021⋅43,47), (a−2021,b−2021)=(472,432) and (a−2021,b−2021)=(2021,2021). Consequently, (a,b)=(2021⋅2022,2022), (a,b)=(2021⋅48,43⋅48), (a,b)=(2021⋅44,47⋅44), (a,b)=(47⋅90,43⋅90), or (a,b)=(2021⋅2,2021⋅2).