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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




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 UTC6, 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 repost 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.



