Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's consider the polynomial $$ f\left( x \right) = \left( {x^2 + 2} \right)\prod\limits_{i = - k}^k {\left( {x - 2i} \right) + 2 \in {\Bbb Q}\left[ x \right]} $$ . Let's suppose that $ p = 2k + 3 \geqslant 5 $ is prime.

Prove the following:

$i)$ Prove that $f$ is irreducible and of degree $p$

$ii)$ Prove that p has exactly $p-2$ real zeros. I have no idea how to prove this, maybe with einsenstein and considering the derivate, but how? :/

share|improve this question
    
Can you prove that $f$ has degree $p$? –  Gerry Myerson Nov 18 '12 at 22:49

1 Answer 1

Modulo $2$, $f(x)$ is just $x^p$; also, the constant term of $f(x)$ is 2; thus, you can apply Eisenstein with $p=2$.

share|improve this answer
    
Ok , and how can I prove that has exactly p-2 real zeros? –  Daniel Nov 18 '12 at 23:01
    
Maybe you can show it changes sign between $n-(1/2)$ and $n+(1/2)$ for $n=-k,\dots,k$. –  Gerry Myerson Nov 18 '12 at 23:10
    
Well It's easy to see the following relation, let's call the polynomial $ f_k $ so we can deduce the following $$ f_k = \left( {f_{k - 1} - 2} \right)\left( {x - 2k} \right)\left( {x + 2k} \right) + 2 $$ so derivating $ f'_k = f'_{k-1} (x-2k)(x+2k) $ so, in each step I add two more critical points, and clearly $ f'_1 = (x^2-2)(5x^2+4) $ thus $ f'_k$ has exactly $2k$ critical points –  Daniel Nov 18 '12 at 23:19
    
Not sure I follow your differentiation --- was expecting the Product Rule to make an appearance. –  Gerry Myerson Nov 19 '12 at 1:12
    
My derivate is bad –  Daniel Nov 20 '12 at 5:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.