Problem 2 of the Canadian Mathematical Olympiad 1978

Find all pairs a, b of positive integers satisfying the equation 2a² = 3b³.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my lovely wife Tran Thi Quynh Chau)

The product on the left side 2a² is even, so 3b³ has to be even, and, therefore, b³ has also to be even, and so has b. Let b = 2n, where n is a positive integer.

We then have b² = 4n². Now rewrite 2a² = 3b³ as 2a² = 3b×4n² which is simplified to

a² = 6bn²

Since a² and n² are already squares of two numbers, 6b has to be the square of another number. Let it be 6b = m² such that b = 6k² where k is a positive integer. Now substitute it to the original equation to obtain

a² = 3b³/2 = 3²×6²(k³)²,

which gives

a = 18k³

The solutions are (a, b) = (18k³, 6k²) where k is a positive integer.

For example, for k = 23, a = 18×23³ = 219006 and b = 6×23² = 3174 is a solution with

		2 x 219006 = 3 x 3174 = 95,927,256,072