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Many triplets II (Posted on 2011-05-07)

Prove that the equation x²+2y²=3z², with gcd(x,y,z) = 1 has an infinite number of positive integer solutions.

See The Solution Submitted by K Sengupta

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Solution

Comment 4 of 4 |

Let y = 1 so that x² + 2y² = 3z² reduces to the Pell-type equation:

                                                x² - 3z² = -2                               (1)

Sqrt(3) has convergents: 1/1, 2/1, 5/3, 7/4, 19/11,..... with alternate values

of (x/z) giving the infinite set of solutions of equation (1) and therefore of the

original equation, as follows:

(x,y,z):      (1,1,1), (5,1,3), (19,1,11), (71,1,41), (265,1,153), (989,1,571),..

Further values can be found using the recurrence relations

            x_n = 4x_n-1 - x_n-2               and       z_n = 4z_n-1 - z_n-2                (2)

This is not as ambitious as Broll’s 2-parameter solution, but we can be sure that all these triplets have gcd(x,y,z) = 1. This is because (2) ensures that all x and z values are odd, and (1) ensures that x and z can therefore have no common prime factor.

Posted by Harry on 2011-05-17 17:20:18

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