Irreducible polynomial Mod $2$ would the polynomial $x^5+x^3+1$ Mod $2$ be irreducible?
I've tried factorizing, however, can't seem to find any factors? if it is irreducible how can i conclude this? 
 A: It's easily checked that $f(x) = x^5 + x^3 + 1$ has no roots over $\mathbb{Z}_2$.  Therefore, if it is reducible, then the only remaining possibility is that it can be written as a product of an irreducible quadratic and an irreducible cubic in $\mathbb{Z}_2[x]$.  
From here, it's pretty straightforward to list every irreducible quadratic and cubic in $\mathbb{Z}_2[x]$ and multiply every possible pair of them.  As a hint in listing these polynomials, keep in mind that a polynomial of degree $\leq 3$ is irreducible $\iff$ it has no roots.  
Even easier: You'll discover that there is only one irreducible quadratic over $\mathbb{Z}_2$.  To save time, you can instead perform polynomial long division to show that it is not a factor of $f$, and thus $f$ is irreducible.
Bottom line: There are only a finite number of possible quadratic/cubic factors of $f$ since we are working over $\mathbb{Z}_2$.  Eliminate all of them, and you can safely conclude that $f$ has no factors.
A: One quick way to prove that $x^5+x^3+1$ is irreducible in $\mathbb Z_2[x]$ is to prove that $x^{32}-x$ is divisible by $x^5+x^3+1$ and that $x^5+x^3+1$ has no linear factors.
More generally, if $f(x)\in\mathbb Z_p[x]$ is of degree $q$, with $q$ prime, then $f(x)$ is irreducible if and only if it is a factor of $x^{p^q}-x$ and it has no linear factors.
You have to switch to cyclomatic polynomials if you want to deal with the case of $q$ not prime. Then, you need $\Phi_{p^q}(x)$ divisible by $f(x)$.
