PUTNAM MATHEMATICS COMPETITION 2018

QUESTION: Find all pairs of natural numbers (a,b) such that:

1/a + 1/b =3/2018

SOLUTION: We start solving the equation:

1/a + 1/b =3/2018

To make it a linear equation, we multiply both the sides of the equation by 2018ab to get

     2018a+2018b=3ab

or, 3ab–2018a–2018b=0

To factorise the LHS, we multiply both sides of the equation by 3 to get

     3.3ab–3.2018a–3.2018b=0

Adding both sides by 2018² and rearranging, we have

     (3a)(3b)–(3a)2018–2018(3b)+2018²=2018²

or, 3a(3b–2018)–2018(3b–2018)=(2×1009)²

or, (3a–2018)(3b–2018)=2²×1009², where 2 and 1009 are prime numbers.

Case1: 3a–2018=1 and 3b–2018=2018²

Solving them,  we get (a,b)=(673,673×2018)

Note that second equation is solved by taking 2018 in RHS

Case2: 3a–2018=2 and 3b–2018=2×1009²

Solving this, we get a=2020/3, which is not a natural number. 

Case3: 3a–2018=2² and 3b–2018=1009²

Solving them, we get (a,b)=(674,337×1009)

Note that second equation is solved by writing 2018 as 2×1009 and taking it in RHS

Case4: 3a–2018=1009 and 3b–2018=2²×1009

Solving them, we get (a,b) =(1009,2018).

Since the equation is symmetric in a and b, so if (a,b) satisfies the equation, then (b,a) also satisfies. So the final solutions are:

(a,b)=(673,673×2018),(674,337×2018),(1009,2018),(673×2018,673),(337×2018,674) and (2018,1009).