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

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Bill Dubuque

Posts: 1,739
Registered: 12/6/04
Re: Another Diophantine equation
Posted: Aug 29, 2011 10:32 PM
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Timothy Murphy <gayleard@eircom.net> wrote:
> José Carlos Santos wrote:
>> On 28-08-2011 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^2-2y^2).
>>>
>>> To make n an integer, simply solve the Pell equation x^2-2y^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



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