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Topic: If a^2 + b^2 – a + b – c = a^2 + b^2, and c is <
ab/3, then a^2 + b^2 is prime

Replies: 8   Last Post: Nov 11, 2017 11:37 PM

 Messages: [ Previous | Next ]
 richard miller Posts: 170 Registered: 9/29/06
Re: If a^2 + b^2 – a + b – c = a^2 + b^2, and c
is < ab/3, then a^2 + b^2 is prime

Posted: Nov 11, 2017 2:52 PM

On Saturday, 11 November 2017 18:55:44 UTC, 4mus...@gmail.com wrote:
> On Saturday, November 11, 2017 at 6:57:35 AM UTC-6, richard miller wrote:
> > On Saturday, 11 November 2017 12:38:26 UTC, 4mus...@gmail.com wrote:
> > > True or False?
> >
> > Are you sure that is what you mean?
> >
> > a^2 + b^2 - a + b - c = a^2 + b^2
> >
> > subtracting a^2 + b^2 from both sides implies
> >
> > - a + b - c = 0
> >
> > rearranging
> >
> > c = b - a
> >
> > Fair enough, perhaps, but a^2 + b^2 can be anything you like.
> >
> > I presume a re-post is on its way?
> >
> > Richard Miller
> > http://www.urmt.org

> OK apparently I crashed and burned, but at least I got a response.
> urmt.org looks great
> I have an idea on Eigenvectors

You've now got two responses.

Mathematics is remorseless, and we all have to suffer its brutal logic.

Richard M.

Date Subject Author
11/11/17 4musatov@gmail.com
11/11/17 richard miller
11/11/17 4musatov@gmail.com
11/11/17 4musatov@gmail.com
11/11/17 4musatov@gmail.com
11/11/17 richard miller
11/11/17 4musatov@gmail.com
11/11/17 Bill
11/11/17 4musatov@gmail.com