Show that $p(x)=x^5+x^3+x+1$ is irreducible.
My attempt: First, I tried to shifting Eisenstein's up to $10$, i.e., $p(x+a)$, where $a=1,\dots, 10$ using calculator. I did not continue because if I am being honest, you will get bored after a several failures. Also, please, I ask you to do not recommend rational root test for irreducibility.
Actually my original question was $x^5+Ax^3+Ax+1$ where $A \in \mathbb Z$.
I put myself this example to observe my own special version, but even can't show easier version. Or, my question can be
Is $x^7+Ax^5+Bx^3+Ax+1$, with $A,B\in\mathbb Z$, irreducible?
(YES! But how can we show?). I couldn't solve also this example.
I know that one way to show is the following: If polynomial is irreducible for some modulo $p$ then it is irreducible over $\mathbb Z$ and by Gauss Lemma irreducible over $\mathbb Q$.
Any help is appreciated.
Best Answer
Working over $\Bbb{F}_2$ it is clear that $1\in\Bbb{F}_2$ is a root of $x^5+x^3+x+1\in\Bbb{F}_2[x]$, and we find that $$x^5+x^3+x+1\equiv(x+1)(x^4+x^3+1)\pmod{2}.$$ The quartic factor clearly has no roots in $\Bbb{F}_2$, and because the only irredicuble quadratic over $\Bbb{F}_2$ is $x^2+x+1$, it suffices to verify that $$x^4+x^3+1\neq(x^2+x+1)^2,$$ to conclude that $x^4+x^3+1$ is irreducible over $\Bbb{F}_2$. Over $\Bbb{Q}$, this means that $x^5+x^3+x+1$ has an irreducible factor of degree at least $4$. That is to say, it is either irreducible or it has a root in $\Bbb{Q}$. The rational root test tells you that if it has a root in $\Bbb{Q}$, then that root must be $1$ or $-1$. But a quick check shows that these are not roots, and hence the original polynomial is irreducible over $\Bbb{Q}$.
The proof of the more general case of $x^5+Ax^3+Ax+1$ can be handled in a similar way, albeit with a bit more work. Of course the argument above holds identically for any odd $A\neq1$. But for even $A$ it does not.
If we try the same argument over $\Bbb{F}_3$ we get the following factorizations into irreducible factors: $$x^5+Ax^3+Ax+1\equiv\begin{cases} (x+1)(x^4+2x^3+x^2+2x+1)&\text{ if } A\equiv0\pmod{3}\\ x^5+x^3+x+1&\text{ if } A\equiv1\pmod{3}\\ (x+2)(x^4+x^3+2)&\text{ if } A\equiv2\pmod{3} \end{cases}$$ So for $A\equiv1\pmod{3}$ we immediately get that $x^5+Ax^3+Ax+1$ is irreducible over $\Bbb{Q}$. For $A\equiv0,2\pmod{3}$ we see, as before, that $x^5+Ax^3+Ax+1$ has an irreducible factor of degree at least $4$. If it has an irreducible factor of degree exactly $4$ then it must also have a linear factor, and hence a rational root. By the rational root test this root must be either $1$ or $-1$, which tells us that either $$1^5+A\cdot1^3+A\cdot1+1=0\qquad\text{ or }\qquad (-1)^5+A\cdot(-1)^3+A\cdot(-1)+1=0,$$ that is to say, either $A=-1$ or $A=0$. Indeed for these two values of $A$ the polynomial $x^5+Ax^3+Ax+1$ is clearly reducible over $\Bbb{Q}$, and then for all other values of $A$ it is irreducible.
As for the septic polynomial $x^7+Ax^5+Bx^3+Ax+1$ with $A,B\in\Bbb{Z}$, it is not irreducible for all $A,B\in\Bbb{Z}$. For example, by the rational root theorem it has a linear factor if and only if $1$ or $-1$ is a root, or equivalently, if and only if $$1^7+A\cdot1^5+B\cdot1^3+A\cdot1+1=0\qquad\text{ or }\qquad (-1)^7+A\cdot(-1)^5+B\cdot(-1)^3+A\cdot(-1)+1=0,$$ which is in turn equivalent to $2A+B=0$ or $2A+B=-2$. There are also values of $A$ and $B$ for which it is reducible but without linear factors. Finding them all is an interesting exercise!