The examples suggest that the choice of sign in the numerator of n_k should be -(-1)^k. It would be a nice result to show that this is the case generally.
Wow, that was hard. Finding one solution was easy. The larger number is odd while the smaller number is even and a bit larger than half the larger, so consider the list of Pythagorean triples to find 15,8,17. Thus n=7 is one of the solutions.
Pell equation!
Very good explanation!
Thank you for sharing!
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Great
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The examples suggest that the choice of sign in the numerator of n_k should be -(-1)^k. It would be a nice result to show that this is the case generally.
Thanks
Wow, that was hard. Finding one solution was easy. The larger number is odd while the smaller number is even and a bit larger than half the larger, so consider the list of Pythagorean triples to find 15,8,17. Thus n=7 is one of the solutions.
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Without solving quadratic equation for root, you can directly reduce it to x^2-d y^2=1
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