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Topic: in FLT, is 1+2B= C unique to 1,4,9? #1372 Correcting Math
Replies: 9   Last Post: Jan 6, 2014 5:15 PM

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James Waldby

Posts: 308
Registered: 1/27/11
Re: in FLT, is 1+2B= C unique to 1,4,9? #1372 Correcting Math
Posted: Jan 6, 2014 4:32 AM
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On Mon, 06 Jan 2014 01:57:18 +0000, Wizard-Of-Oz wrote:
> Archimedes Plutonium <plutonium.archimedes@gmail.com> wrote in
> news:c51278d2-8cd2-4b41-b753-9994b5926c76@googlegroups.com:

>>
>> in FLT, is 1+2B= C unique to 1,4,9? #1372 Correcting Math
>>
>> Alright, before I go any further, I need to show or prove that 1+2B= C
>> exists uniquely for exp2 in the case of 1,4,9 and for which it
>> immediately gives a Pythagorean triple of 3,4,5, or A+B=C.

>
> So you want to know if there are two perfect squares (other
> than 4 and 9) where
>
> 1 + 2B = C
>
> so you want natural number solutions for 1 + 2j^2 = k^2
> other than j=2 and k=3


There are infinitely many such solutions; expressing them in
the form k/j [like the high-side approximants to sqrt(2) that
are every other term in the continued fraction for sqrt(2)]
they include 17/12, 99/70, 577/408, 3363/2378, etc.

For example, 1 + 2*70^2 = 1+2*4900 = 9801 = 99^2.

The low-side approximants, like 7/5, 41/29, 239/169, etc.,
solve the equation -1 + 2*j^2 = k^2. Eg, -1 + 2*25 = 49.

--
jiw



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