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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 ]
 4musatov@gmail.com Posts: 199 Registered: 10/2/17
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 1:55 PM

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

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