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**NonCommAlg** well, $\displaystyle y^2+y+ \nu$ has degree 2. so it's irreducible iff it has no root in $\displaystyle \text{GF}(16).$ thus the polynomial is irreducible for all odd values of $\displaystyle \nu$ in $\displaystyle \text{GF}(16)$ because $\displaystyle y^2+y$ is always even. so the only thing

you need to prove is that the polynomial has a root for 8 remaining even values of $\displaystyle \nu.$ this is clear because in $\displaystyle \text{GF}(16): \ \{y^2+y: \ 0 \leq y \leq 7 \}=\{0,2,4,6,8,10,12,14 \}.$

__A Much Better Question__: given a prime number $\displaystyle p$ and $\displaystyle n \in \mathbb{N},$ find the number of elements of $\displaystyle \nu$ in $\displaystyle \text{GF}(p^n)$ for which the polynomial $\displaystyle y^2+y+\nu$ is irreducible.