Cool polynomial forms

Let \(P(x)\) be a given polynomial with integer co-efficients.Prove that there exists polynomials \(Q(x)\) and \(R(x)\) again with integer co-efficients such that

(a) \(P(x)Q(x)\) is a polynomial in \(x^{2}\)

(b) \(P(x)R(x)\) is a polynomial in \(x^{3}\)

Note by Eddie The Head
4 years, 1 month ago

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(a) \( Q(x) = P(-x) \). (b) \( R(x) = P(\omega x) P(\omega^2 x) \), where \( \omega = {\rm exp}(2\pi i / 3). \)

In general, if \( Q(x) = \prod_{k=1}^{n-1} P(\zeta_n^k x) \), then \( P(x)Q(x) \) is a polynomial in \( x^n \). It shouldn't be too hard to see that \( Q(x) \) has integer coefficients.

Maybe the easiest way to prove these statements is to factor \( P(x) \) over the complex numbers and then to use that \( (x-\alpha) \prod_{k=1}^{n-1} (x-\zeta_n^k \alpha) = x^n-\alpha^n \).

I will let you all fill in the details. :)

Patrick Corn - 4 years, 1 month ago

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Can you explain why would the polynomial would have integer coefficients?

Calvin Lin Staff - 4 years, 1 month ago

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Fine, fine, call my bluff why don't you!

Well, it should have rational coefficients by Galois theory (or whatever), since the polynomial's symmetry implies that it's fixed by all the automorphisms of the complex numbers, because they permute the nontrivial \( n \)th roots of unity.

But the coefficients are also in \( {\mathbb Z}[\zeta_n] \), and it's well-known that the intersection of \( {\mathbb Z}[\zeta_n] \) and \( \mathbb Q \) is just \( \mathbb Z \): put plainly, they must be integers since they're algebraic integers.

I think that works.

Patrick Corn - 4 years, 1 month ago

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