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Many triplets II (Posted on 2011-05-07) Difficulty: 3 of 5
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    
Rating: 2.5000 (2 votes)

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Solution 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

            xn = 4xn-1 - xn-2               and       zn = 4zn-1 - zn-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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