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Re: in FLT, is 1+2B= C unique to 1,4,9? #1372 Correcting Math
Posted:
Jan 6, 2014 4:32 AM


On Mon, 06 Jan 2014 01:57:18 +0000, WizardOfOz wrote: > Archimedes Plutonium <plutonium.archimedes@gmail.com> wrote in > news:c51278d28cd24b41b7539994b5926c76@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 highside 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 lowside approximants, like 7/5, 41/29, 239/169, etc., solve the equation 1 + 2*j^2 = k^2. Eg, 1 + 2*25 = 49.
 jiw



