On Saturday, May 6, 2017 at 2:30:15 PM UTC+3, bassam king karzeddin wrote:
> > On 2017-05-03 18:00:01 +0000, quasi said:
> >
> > > Earle Jones27 wrote:
> > >>
> > >> Let's think of a Pythagorean triple (like 3, 4, 5)
> > of the form:
> > >>
> > >> a^2 + b^2 = c^2
> > >>
> > >> a, b, c integers.
> > >>
> > >> b = a+1
> > >>
> > >> and....
> > >>
> > >> c is a perfect square.
> > >>
> > >> So that we could write
> > >>
> > >> a^2 + b^2 = d^4
> > >>
> > >> To get you started:
> > >>
> > >> Try a = 119
> > >> b = 120
> > >> c = 169 ( = 13^2)
> > >>
> > >> Others?
> > >
> > > If there's another such triple, then a will have
> > more
> > > than 10^7 decimal digits.
> > >
> > > quasi
> >
> > *
> > Ever wonder why our resident geniuses (John Gabriel,
> > Archimedes
> > Plutonium, Pentcho Valev, Bassam King Karzeddin and
> > others) do not
> > chime in on this problem, which should be very simple
> > for the truly
> > outstanding minds?
> >
> > More on this problem later.
> >
> > earle
> > *
> >
> This also is a Pythagorean triplet that might help
>
> (a^4 - (2b^2)^2), (2ab)^2, (a^4 + (2b^2)^2)
>
> Where (a, b) are non-zero integers, and If you need it in primitive form, then (a, b) must be positive coprime integers with different polarity, (i.e odd and even)
>
> Let me guess that you mean I am a crank by listing me with others whom they were called so by the common publics, but even though those who make distinguished claims based on some logic are far better than those who are blind followers
>
> And Frankly, and unlike so many others, you can not simply ignore my irrefutable (formulas, conjectures, proved theorems, puzzles, ..., etc)
>
> They are still standing there in my posts, wonder!
>
> But it seems it is too embarrassing to challenge them, so better keep them away, for sure
>
> If you need more help, just ask for it
>
> Regards
> Bassam King Karzeddin
> 6 May 2017