by stupidityismygam » Thu May 06, 2010 11:42 am
You are overshooting by about a factor of 3.
We have that [unparseable or potentially dangerous latex formula] and that [unparseable or potentially dangerous latex formula]. Given that the roots of [unparseable or potentially dangerous latex formula] are [unparseable or potentially dangerous latex formula] with [unparseable or potentially dangerous latex formula], it is possible to generate 3 other distinct polynomials with roots of [unparseable or potentially dangerous latex formula]. However, if [unparseable or potentially dangerous latex formula] then it is only possible to generate 2 other distinct polynomials with roots of [unparseable or potentially dangerous latex formula]. So we can simply assume, for now, that [unparseable or potentially dangerous latex formula] and then account for the other polynomials generated.
From Vieta, we have that [unparseable or potentially dangerous latex formula] and since [unparseable or potentially dangerous latex formula] we have that there is a bijection between the number of polynomials with roots [unparseable or potentially dangerous latex formula] and the factors of [unparseable or potentially dangerous latex formula] divided by 2 (assuming that [unparseable or potentially dangerous latex formula]). So, we are simply looking for the sum of the number of factors of the factors of 2010. This is simply [unparseable or potentially dangerous latex formula]. Note that when [unparseable or potentially dangerous latex formula], it produces a double root and so the number of polynomials [unparseable or potentially dangerous latex formula]
my avatar is pretty awesome
tytia