Estonian Math Competitions

Solution 1: Let $d=\gcd (a,b)$, $a=d{a}^{\prime }$, $b=d{b}^{\prime }$. The given equation reduces to $\frac{1}{d{a}^{\prime }}+\frac{1}{d{b}^{\prime }}=\frac{1}{2021}$ which is equivalent to $$2021(a^{\prime }+b^{\prime })=d{a}^{\prime }{b}^{\prime }.(4)$$ As $a^{\prime }$ and $b^{\prime }$ are relatively prime to each other, they both are relatively prime to $a^{\prime }+b^{\prime }$, implying that $a^{\prime }{b}^{\prime }$ and $a^{\prime }+b^{\prime }$ are relatively prime. By (4), $a^{\prime }{b}^{\prime }\mid 2021(a^{\prime }+b^{\prime })$, implying $a^{\prime }{b}^{\prime }\mid 2021$. As $2021=43\cdot 47$ where the factors are prime, the number $2021$ has exactly four positive factors $1$, $43$, $47$ and $2021$. Taking into account that $a\ge b$ holds if and only if $a^{\prime }\ge b^{\prime }$, consider all cases: If $a^{\prime }=2021$ and $b^{\prime }=1$ then (4) implies $d=2022$. Consequently, $(a,b)=(2021\cdot 2022,2022)$. If $a^{\prime }=47$ and $b^{\prime }=43$ then (4) implies $d=90$. Consequently, $(a,b)=(47\cdot 90,43\cdot 90)$. If $a^{\prime }=47$ and $b^{\prime }=1$ then (4) implies $d=43\cdot 48$. Consequently, $(a,b)=(2021\cdot 48,43\cdot 48)$. If $a^{\prime }=43$ and $b^{\prime }=1$ then (4) implies $d=47\cdot 44$. Consequently, $(a,b)=(2021\cdot 44,47\cdot 44)$. If $a^{\prime }=1$ and $b^{\prime }=1$ then (4) implies $d=2021\cdot 2$. Thus $(a,b)=(2021\cdot 2,2021\cdot 2)$.

Solution 2:* The given equation is equivalent to $\frac{a+b}{ab}=\frac{1}{2021}$ which reduces to $ab-2021a-2021b=0$. After adding $2021^2$ to both sides and factorizing in the l.h.s, we get $$(a-2021)(b-2021)=2021^2.(5)$$ 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 $2021^2$ and (5) could not hold. Hence both factors are positive. Since $a\ge b$, we must have $a-2021\ge b-2021$. As $2021=43\cdot 47$ with factors being prime, we obtain variants $(a-2021,b-2021)=(2021^2,1)$, $(a-2021,b-2021)=(2021\cdot 47,43)$, $(a-2021,b-2021)=(2021\cdot 43,47)$, $(a-2021,b-2021)=(47^2,43^2)$ and $(a-2021,b-2021)=(2021,2021)$. Consequently, $(a,b)=(2021\cdot 2022,2022)$, $(a,b)=(2021\cdot 48,43\cdot 48)$, $(a,b)=(2021\cdot 44,47\cdot 44)$, $(a,b)=(47\cdot 90,43\cdot 90)$, or $(a,b)=(2021\cdot 2,2021\cdot 2)$.