(a) $x^2+x+1∈\mathbb{F}_2[x]$
(b) $x^3+x+1∈\mathbb{F}_3[x]$
(c) $x^4+1∈\mathbb{F}_5[x]$
(d) $x^4+10x+1∈\mathbb{Z}[x]$
2. Prove the following polynomials are irreducible in $\mathbb{Z}[x]$:
(a) $x^4-4x^4+6$
(b) $x^6+30x^5-15x^3+6x-120$
(c) $x^4+4x^3+6x^2+2x+1$
(d) $\dfrac{(x+2)^p-2^p}{x}$ for $p$ an odd prime.
3. Show that $f_n(x)=(x-1)(x-2)...(x-n)-1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$.
4. Show that $f_n(x)=(x-1)(x-2)...(x-n)+1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$ and $n≠4$.
Proof: (1) (a) There are no zeros to this equation, so it is irreducible.
(b) By discovering $f(1)=0$, we obtain $x^3+x+1=(x-1)(x^2+x+2)$, where the latter has no zeros.
(c) $x^4+1=x^4-4=(x^2-2)(x^2+2)$.
(d) Reduce modulo 5. If $f(x)=a(x)b(x)$, then due to monicity of $f(x)$ we have $a(x)$ and $b(x)$ are monic and not collapsing to units so that $\overline{a(x)}$ is associate to $\overline{x^2+2}$ or $\overline{x^2-2}$, say the former. Since $\overline{a(x)}$ is monic we have $\overline{a(x)}=\overline{x^2+2}$ which forces the constant term of $a(x)$ to not be $\pm 1$, a contradiction.
Alternatively, since $f(x)$ clearly has no zeros in $\mathbb{Z}$, it would have to factor as the product of two quadratics such that the following hold:$$f(x)=a(x)b(x)$$$$a(x)=x^2+a_1x+a_0$$$$b(x)=x^2+b_1x+b_0$$$$a_0b_0=1$$$$a_1b_0+a_0b_1=0$$$$a_1b_1+a_0+b_0=10$$$$a_1+b_1=0$$Solving for $b_1$ as a typical system of equations reveals$$\dfrac{1}{4}b_1^4+15b_1^2+24=0$$which is impossible.
(2) (a) Eisenstein's criterion for $p=2$ applies.
(b) Eisenstein applies for $p=3$.
(c) We generalize the technique employed in example 2. Lemma 1: Let $R$ be an integral domain with $f(x)=a_0+a_1x+...+a_nx^n∈R[x]$ and $g(x)=f(r(x))$ for $r(x)$ a nonconstant polynomial. If $g(x)$ is irreducible, then $f(x)$ is irreducible. Proof: If $f(x)=a(x)b(x)$ for some nonunit polynomials, then $g(x)=a(r(x))b(r(x))$ where necessarily $\text{deg}(a(r(x)))=\text{deg}(a(x)) \cdot \text{deg}(r(x)) \geq 1$ if $\text{deg}(a(x)) \geq 1$ and $a(r(x))=a(x)$ if $a(x)$ is a constant and similarly for $b(r(x))$, a contradiction.$~\square$
Let $g(x)=f(x-1)=x^4-2x+2$, where Eisenstein applies for $p=2$.
(d) $\dfrac{(x+2)^p-2^p}{x}=x^p+...+2p$, where by the binomial theorem the omitted terms have coefficients of multiples of $p$. Eisenstein applies for $p$.
(3) When $n=1$ we clearly have $f_n(x)$ is irreducible, so assume $n > 1$. Assuming $f_n(x)$ has a zero in $\mathbb{Z}$, then $1=(x-1)(x-2)...(x-n)$ so that in particular $x-1=\pm 1$ and $x-2=\pm 1$, a contradiction. Therefore if $f_n(x)=a(x)b(x)$ then the degrees of $a(x)$ and $b(x)$ are greater than one.
We demonstrate a basic algebraic result, first. If $R$ is an integral domain and $f(x)∈R[x]$ of degree $n \geq 1$, then there are at most $n$ zeros of $f(x)$. If not, then $f(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n+1})f'(x)$ entails $\text{deg}(f(x)) > n$, a contradiction.
Since $f_n(m)=-1$ for all integers $1≤m≤n$ we have $a(m)= \pm 1$ and $b(m)= \mp 1$ for at least $n$ values. We thus have $a(x)+b(x)$ has at least $n$ zeros, so that necessarily $a(x)=-b(x)$. But now the largest term's coefficient of $a(x)b(x)$ is $-1$, a contradiction.
(4) Similar as in the last problem, $f_1(x)$ is irreducible and $f_n(x)$ has no zeros for $n > 1$, so that necessarily $a(x)=b(x)=\pm 1$ for at least $n$ values, and again $a(x)=b(x)$ so that $f_n(x)$ is a perfect square. However, when $n$ is odd this is an impossibility by degree, so $n$ must be even. For $n < 4$ we can observe $f_n(4)$ is irreducible, and for $n > 4$ and $n$ even we can observe $f_n(3/2)$ is negative, a contradiction since $f_n(x)$ is a perfect square.
When $n=4$, with a process similar to that for (1)(d), we can observe $f_4(x)=(x^2-5x+5)^2$ is its decomposition.$~\square$
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