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Posts Tagged ‘irreductible’

## Irreductible polynomial with paired roots

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 $f$.

## Irreductible Polynomial TST 2003

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

Categories: IMO, Problem Solving Tags: ,

## Irreductible polynomial

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

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

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 $\mathbb{Z}[X]$.

## Switch the coefficients

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

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 $n>m>0$.
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 $\Bbb{Q}$.