Show that the polynomial $7X^5 + 71X^3 - 9$ is irreducible in $\mathbb{Z}[X]$ Show that the polynomial $f(X) = 7X^5 + 71X^3 - 9$ is irreducible in $\mathbb{Z}[X]$
My solution:
Using the irreducibility test: "Reduction Mod p Test"
$f(X)$ is clearly primitive and that the prime number 2 does not divide the leading coefficient.
It is therefore enough to prove the polynomial $\bar{\pi}_2(f(X)) = X^5+X^3-\bar{1} \in (\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible.
$\bar{\pi}_2(f(X))$ has no roots in $(\mathbb{Z}/2\mathbb{Z})[X]$ (to verify this evaluate the polynomial at the two elements $\bar{0}$ and $\bar{1}$ of $(\mathbb{Z}/2\mathbb{Z})[X]$
Thus, if it were reducible then it would have to be of the
form $g(X) \cdot h(X)$, where $g(X), h(X) \in (\mathbb{Z}/2\mathbb{Z})[X]$ are both irreducible of degree $2$. Which is not possible.
Concluding that $\bar{\pi}_2(f(X))$ is irreducible in $(\mathbb{Z}/2\mathbb{Z})[X]$, whereby $f(X)$ is irreducible in $\mathbb{Z}[X]$.
 A: Modulo $2$, this polynomial is nothing else than $X^5+X^3+1$. As it has no root in $\mathbf F_2$, if it could be factored in $\mathbf F_2[X]$, it would the product of an irreducible quadratic factor and an irreducible cubic  factor. Now there's only one irreducible quadratic polynomial in  $\mathbf F_2[X]$: $\;X^2+X+1$, and you can check the result of the Euclidean division of $X^5+X^3+1$ by $X^2+X+1$ has a remainder of $X+1$.
A: $\bar f(x)=x^5+x^3+1$ would have to be divisible by $x^2+x+1$, since the other three degree two polynomials are reducible. 
So suppose $\bar f(x)=g(x)(x^2+x+1)$.  Then let $g(x)=x^3+ax^2+bx+c$.  Then $\bar f(x)=x^5+(a+1)x^4+(1+a+b)x^3+(a+b+c)x^2+(b+c)x+c$.  So $a=1, c=1,b=1$ which implies $a=0$, a contradiction. 
That is, the system of equations is inconsistent,  and $f(x)=7x^5+71x^3-9$ is irreducible. 
A: Hint $f\,$ has no roots in $\Bbb F_2$  so no linear factors so if $f$ splits it has an irred. quadratic factor $g$, so in  $\,\Bbb F_2[x]/g \cong \Bbb F_4\!:\,$ $\,\color{#c00}{x^3\! =\! 1}\,$ so  $\ \color{#0a0}0 \!=\! f\! =\! \color{#0a0}{x^2}(\color{#c00}{x^3})+\color{#c00}{x^3}\!-1 = \color{#0a0}{x^2}\,$ so $\,\color{#c00}{1\! =\! x^3}\! =\! \color{#0a0}{x^2}\color{c00}{x} \!=\!\color{#0a0}0,\,$ contradiction.
Remark $ $ We used $\,0\neq f\in \Bbb F_4\Rightarrow f^{\large \color{#c00}3}\! = 1\,$ (for $\,f\!=\!x),\,$ an analogue of Fermat's little Theorem, which is true because $\,\Bbb  F_4$ has multiplicative group $\,\Bbb F_4^*$ of size $\,\color{#c00}3 =4\color{darkorange}{-1} \,$ ($\rm non\color{darkorange}{zero}$ elements are invertible in a field), so Lagrange's Theorem $\Rightarrow f^{\large\color{#c00}3}\! = 1\,$ for all $\,f\neq 0.$
Above is essentially a special case of a general  polynomial irreducibility test over finite fields - which is an an efficient analog of the impractical Pocklington-Lehmer integer primality test.
