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 $4(n+1)^2 + 15(n+1) - 1 = 4n^2 + 23n + 18 = 9k + 8n + 19$
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1Hello. What did you try and where did you get stuck? – Mankind Aug 20 '15 at 20:52
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7I think this is false, since for $n=3$ you get 80. – Rolf Hoyer Aug 20 '15 at 20:56
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@RolfHoyer You're right :(. I struggled with this for 1 hour and it seems like my math textbook is wrong. – user4201961 Aug 20 '15 at 20:58
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I think he meant $4^n +15n -1$ is divisible by 9. – Anonymous196 Sep 26 '17 at 14:07
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Maybe you meant this https://math.stackexchange.com/questions/1404305/9-mid-4n2-15n-1-for-n-in-mathbb-n?rq=1 – Anonymous196 Sep 26 '17 at 14:48
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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.
egreg
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It is not true buddy, take $n=3$. In general if $n=3t$ it wont be true. Greetings!