×

# 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 $$x^{3}$$

3 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

(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. :)

- 3 years, 8 months ago

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

Staff - 3 years, 8 months ago

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.

- 3 years, 8 months ago