Re: Polynomials of degree 2 with odd coefficients
- From: Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx>
- Date: 09 May 2006 03:03:37 -0400
JoeS <jhs@xxxxxxxxxxxxxx> wrote: (*edited*)
Ana wrote: (*paraphrased*)
Show P(x) = ax^2+bx+c has no rational roots if a,b,c are all odd
Let r = n/m where the fraction n/m is in lowest terms.
Then m^2 P(n/m) = a n^2 + b nm + c m^2
Consider the three cases: n,m both odd; n odd,m even; n even,m odd
In each case the integer m^2 P(n/m) is odd, so it can't be zero.
This holds true for any polynomial - as I posted a few years ago:
THEOREM P in Z[x] has a root in Q -> P(0) P(1) P(oo) is even,
i.e. odd leading coef P(oo) -> const coef or coef sum is even
PROOF P(n/m)=0, gcd(n,m)=1, odd leading coef -> m odd, by
Rational Root Test [2], so Mod 2: m=1 -> P(n) = P(n/m) = 0. QED
It is a monic version of the PARITY ROOT TEST in my 2003/3/3 post:
http://google.com/groups?selm=y8zof3ohyb4.fsf%40nestle.ai.mit.edu
--Bill Dubuque
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