<s>[INST] $9 \mid 4n^2 + 15n - 1$ for $n \in \mathbb N$ How to prove by induction that $9 \mid 4n^2 + 15n - 1$ for every $n \in \mathbb N$?
For $n = 1$
$4 \cdot 1^2 + 15 \cdot 1 - 1 = 18$
For $n \ge 2$
If $4n^2 + 15n - 1 = 9k$ then $4(n+1)^2 + 15(n+1) - 1 = 4n^2 + 23n + 18 = 9k + 8n + 19$
 [/INST] If $4n^2+15n-1$ is divisible by $9$, then it's also divisible by $3$; however
$$
4n^2+15n-1\equiv n^2-1\pmod{3}
$$
but $n^2\equiv 1\pmod{3}$ if and only if $3\nmid n$. So, for $n=3k$, the number $4n^2+15n-1$ is not divisible by $3$ and, of course, not divisible by $9$ either.
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