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.
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.