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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?

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    $\begingroup$ It is general fact that a prime $p$ is $\equiv 3\pmod{4}$ iff there is no $x\in \mathbb{Z}$ such that $x^2 \equiv -1\pmod{p}$. $\endgroup$ – Prahlad Vaidyanathan Nov 28 '13 at 16:18
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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.

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