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Posts Tagged ‘irreductible’

Irreductible polynomial with paired roots

April 11, 2011 Leave a comment

Suppose f \in \Bbb{Q}[X] is an irreductible polynomial which has a complex root a such that -a is also a root for f. Prove that for any other root b of f, -b is also a root for f.

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Irreductible Polynomial TST 2003

December 28, 2010 Leave a comment

Let f\in\mathbb{Z}[X] be an irreducible polynomial over the ring of integer polynomials, such that |f(0)| is not a perfect square. Prove that if the leading coefficient of f is 1 (the coefficient of the term having the highest degree in f) then f(X^2) is also irreducible in the ring of integer polynomials.

Mihai Piticari, Romanian TST 2003

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Categories: IMO, Problem Solving Tags: ,

Irreductible polynomial

January 17, 2010 Leave a comment

Suppose f=X^p+a_{p-1}X^{p-1}+...a_1X+p is a polynomial with integer coefficients such that p is an odd prime, and the roots of f form a regular polygon in the complex plane. Prove that f is irreductible in \mathbb{Q}[X].
Ioan Baetu, Romanian Mathematical Gazette

Irreductible polynomials

November 19, 2009 Leave a comment

1) Prove that f=(x-a_1)(x-a_2)...(x-a_n)-1 is irreductible over \mathbb{Z}[X], where a_1,...,a_n are pairwise different.
2) Prove that f=(x-a_1)(x-a_2)...(x-a_n)+1 is irreductible over \mathbb{Z}[X], where a_1,...,a_n are pairwise different with one single exception, namely, (x-a)(x-a-1)(x-a-2)(x-a-3)+1.
3) Prove that f=(x-a_1)^2(x-a_2)^2...(x-a_n)^2-1 is irreductible over \mathbb{Z}[X], where a_1,...,a_n are pairwise different.
4) Prove that if a polynomial of degree n has integer coefficients and takes the values \pm 1 for more than 2 \cdot\lfloor  \frac{n}{2} \rfloor then this polynomial is irreductible. ( \lfloor x \rfloor is the greater integer not exceeding x )

Irreductible polynomial 1

November 9, 2009 Leave a comment

Prove that if a \in \mathbb{Z}^* and n\geq 2 then the polynomial f(x)=x^n+ax^{n-1}+...+ax-1 is irreductible in \mathbb{Z}[X].
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Switch the coefficients

November 9, 2009 Leave a comment

Let f a polynomial of degree n having integer coefficients; then f is irreductible in \mathbb{Z}[X] if and only if g(x)=x^nf(\frac{1}{x}) is irreductible in \mathbb{Z}[X].
Solution: Suppose f is irreductible and g=pq, with p,q \in \mathbb{Z}[X]. Then f(x)=x^n p(\frac{1}{x})q(\frac{1}{x})=p_1(x)q_1(x) when p_1,q_1 \in \mathbb{Z}[X] are two polynomials of degree at least 1. Therefore f is irreductible. \Box

Categories: Algebra, Problem Solving Tags:

Irreductible Polynomial

November 2, 2009 4 comments

Prove that the polynomial \displaystyle f(x)=\frac{x^n+x^m-2}{x^{\gcd (m,n)}-1} is irreductible over \mathbb Q for all integers n>m>0.
Miklos Schweitzer 2009 Problem 4
A generalization of this problem was proposed by myself in the Romanian TST 2010, the solution being similar to the one below. The generalization said to prove that if p is a prime and k_1,...,k_p are positive integers with d=\gcd (k_1,...,k_p) then \displaystyle f(x)=\frac{x^{k_1}+...+x^{k_n}-p}{x^d-1} is irreductible over \Bbb{Q}.
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