[seqfan] Re: near multiples of squares in Fibonacci, Lucas, Pell, etc. numbers
Max Alekseyev
maxale at gmail.com
Wed Jun 5 12:21:31 CEST 2013
The answer is No.
Notice that a * x^2 + b * x + c = 1/4 * ( (2ax + b)^2 + 4c - b^2 ) = A
* X^2 + B, where A and B are rational coefficients and X = 2ax+b.
I did not make it clear but theorem 7 (and theorem 5 it relies on)
works for rational coefficients and argument as well (via multiplying
equation/form by a suitable integer), implying that there could be
only a finite number of Fibonacci numbers of this form.
I will state this explicitly in the next revision.
Regards,
Max
On Tue, Jun 4, 2013 at 9:16 PM, Allan Wechsler <acwacw at gmail.com> wrote:
> Are there _any_ quadratic polynomials that take on an infinite number of
> Fibonacci values for integral argument?
>
>
> On Tue, Jun 4, 2013 at 9:06 PM, Max Alekseyev <maxale at gmail.com> wrote:
>
>> SeqFans,
>>
>> I've just published at arxiv a manuscript on finding integral points
>> on biquadratic curves with application to finding terms of the form
>> a*m^2 + b in Lucas sequences:
>> http://arxiv.org/abs/1306.0883
>> As an example I established that among Fibonacci numbers only 2 and 34
>> are of the form 2m^2+2; only 1, 13, and 1597 are of the form m^2-3;
>> and so on.
>>
>> I would appreciate any comments or suggestions.
>>
>> P.S. I'm going to present this result at CanaDAM 2013 conference next week.
>>
>> Regards,
>> Max
>>
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>>
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>>
>
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