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Bài giải cho phương trình 2a² = 3b³

[ Friday, 29 January, 2010 ]

Giải một phương trình có hai nghiệm số nguyên dương (positive integers) a và b sao cho 2a² = 3b³.

Nguyên văn là bài thi giỏi toán Canada 1978 như sau. Để bản tiếng Anh cho sinh viên nuớc ngoài search.

Problem 2 of the Canadian Mathematical Olympiad 1978

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

Solution by Vo Duc Dien

The product on the left side 2a² is an even number, so 3b³ has to be an even number,

and b³, therefore, has to be an even number, or b to be an even number. Let b = 2n

where n is a positive integer.

We then have         b² = 4n²       now rewrite 2a² = 3b³  as  2a² / b² = 3b              or

2a² / (4n²) = 3b      or      a² / (2n²) = 3b                 or      a²  = 6bn²

Since a² is the square of a number and n² is already a square of a number, we have

to have  6b   to be the square of another number. Let it be 6b = m²  or

b = 6k²  where k is a positive integer. Now substitute it to the original equation,

we have                 a² = 3b³ /2 = 3²x6²(k³)²   or    a  = 18k³

Solutions are (a, b) = ( 18k³, 6k² )  where k is a positive integer. For example, for

k = 23,  a = 18 x 23³ = 219006 and b = 6 x 23² = 3174 is a set of solution when

                   2 x 219006² = 3 x 3174³ = 95,927,256,072

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