The exercise is given by
In $\mathbb{F}_{743}[x]$ show that $x^2+1$ is irreducible and determine a generator of the field $\mathbb{F}_{743}[x]/(x^2+1)$.
Hmm I'd like to try this by assumption but I don't know how to start:
Assume $\exists k(x),l(x) , \deg(k)=\deg(l)=1$ such that $k(x)*l(l)=x^2+1$ but I don't know how to continue?
On the other hand Wolfram Alpha says there is no solution of $x^2=-1$ in $\mathbb{Z}_{743}$.
Can please someone of you give me a hint how to solve this?
As you say if it was reducible there should be two polynomial of degree equal to $1$, so this polynomial should have at least one solution.
Let be $a\in F_{743}$ such a solution, so $a^2=-1$ and $a^4 =1$, but $4$ does not divide $742 = |F_{743}^*|$. So it is absurd.