[tex]n^2-6n+34 \Leftrightarrow n^2-6n+9+25\\
n^2-6n+9+25=(n-3)^2 + 25\\
(n-3)^2+25 \in \mathbb{N} \Leftrightarrow (n-3)^2 \in \mathbb{N}\\
(n-3)^2 \geq 0 | \surd \\
|n-3| \geq 0\\
n-3 \geq 0 \Rightarrow n \geq 3 \Rightarrow n \in [3,+\infty)\\
3-n \geq 0 \Rightarrow n \leq 3 \Rightarrow n \in (-\infty,3] \\
Atunci \; n \in [3,+\infty) \cup (-\infty,3] \Rightarrow n \in (-\infty,+\infty)\\[/tex]
Deci n poate lua orice valoare în [tex]\mathbb{Z}[/tex]
