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Topic:
Another Diophantine equation
Replies:
7
Last Post:
Aug 30, 2011 8:02 PM




Re: Another Diophantine equation
Posted:
Aug 29, 2011 10:32 PM


Timothy Murphy <gayleard@eircom.net> wrote: > José Carlos Santos wrote: >> On 28082011 21:37, TPiezas wrote: >> >>>> Which numbers of the form 2n^2 + n + 1 are perfect squares? I was able >>>> to check that, among those values of _n_ whose absolute values isn't >>>> greater than 1000, those with this property are 264, 95, 8, 3, 0, >>>> 1, 16, 45, and 552. Is the set of these numbers finite or not? >>> >>> The set is infinite. To solve, >>> >>> 2n^2+n+1 = z^2 >>> >>> Let n = (2x+y)y/(x^22y^2). >>> >>> To make n an integer, simply solve the Pell equation x^22y^2 = +/1 >> >> Thanks a lot. I would have never thought of that. > > While I agree with TPiezas that the secret is > to translate the given equation > 2n^2 + n + 1 = m^2 > into the form of Pell's equation, > it wasn't entirely clear to me how this was done.
Lagrange showed how to do this a couple hundred years ago, see http://math.stackexchange.com/questions/9269/9288#9288
Bill Dubuque



