firstname.lastname@example.org (Andrzej Kolowski) wrote in message news:<200409270403.i8R43fY08660@proapp.mathforum.org>... > On 26 Sep 2004, James Harris wrote: > >I've given the example of > > > >x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > > > >where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer > > > >to highlight the oddity that if s_1 and s_2 are rational, then > >trivially you can have both be units as both can equal 1, but if s_1 > >and s_2 are irrational, then they cannot be units in the ring of > >algebraic integers. > > > >It gets weirder. > > > >I picked 2 and 3 because they're the first two primes--and therefore > >coprime to each other--and are small, but let's replace them with 'a' > >and b, non-unit coprime algebraic integers: > > > > > >x^2 + (as_1 + bs_2)x + s_1 s_2 ab = (x + as_1))(x + bs_2) > > > >where s_1 s_2 = 1, as_1 + bs_2 is an integer, and ab is an integer. > > > >Question: Can s_1 and s_2 be irrational and algebraic integers? > > > > Consider P(x) = x^2 + 55*x + 110. > > The roots are r1, r2 = (-55 +/- sqrt(3025 - 12100))/ 2. > > = 55*(-1 +/- sqrt(-3))/2. > > Letting a = 5 and b = 11, we have > > 5*s1 = -55*(-1 + sqrt(-3))/2 and > > 11*s2 = -55*(-1 - sqrt(-3))/2. > > That is, > > s1 = -11*(-1 + sqrt(-3))/2 and > > s2 = -5*(-1 - sqrt(-3))/2. > > Finally, note that (-1 +/- sqrt(-3))/2 are algebraic > integers. > > Therefore s1 and s2 are irrational and are algebraic > integers. > > But so what? Did you think this couldn't happen?? > > Andrzej >
I have to admit as James Harris pointed out that I overlooked the requirement that s1 * s2 = 1, both in this note and the subsequent one in which
P(x) = x^2 + 100*x + 100.
I would *guess* with that additional restriction, there are not going to be any examples. But again, so what? Why should there be? Is every equation supposed to have a solution in algebraic integers?
Incidentally, the current version of APF still has an error in the last section. There should be an f^2 in the third part of the coefficient of x^3.