Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5? I make some test using python and I find that this is the case for $n = 5k +p$ with $k$ an integer and $p =$1,2,or 3.
I was able to prove for $p = 1$ and $p=2$ but not for $p = 3$.
What I'm doing wrong? This is my work:
Let's assume 5 is a divisor of $8n^3+12n^2-2n-3$. Let's prove it for $n + 1.$
We have $8(n+1)^3+12(n+1)^2-2(n+1)-3=8(n^3+3 n^2+3n+1)+12(n^2+2n+1)-2(n+1)-3=(8n^3+12n^2-2-3)+24n^2+48n+18$
For induction we have $8n^3+12n^2-2n-3$, we need only that $5$ divide $24n^2+48n+18$
Since  $n = 5k + p$, $k$ an integer and $p$ with values $1$,$2$ or $3$.
We need to see that $5$ divide $24n^2+48n+18= 24$ with $(5k+p)^2+48(5k+p)+18$, 
Using congruence we have $(5k+p)^2$ is congruent with $p^2$ mod 5 and $5k + p$ with $p$ modulo 5.
$f(p)= 24p^2+48p+18$ is divisible with 5 para $p = 1, p = 2$.
So $f(1) = 90$ is divisible for 5 , $f(2)= 210$ and also divisible for 5.
But $f(3)$ is not divisible for 5.

OP changed the question:
So, another solution:
$$8n^3+12n^2-2n-3 \equiv_5 3n^2 + 2n^2 -2 n - 3 \stackrel{n \equiv_5 k}{\equiv_5} 3k^3 + 2k^2 - 2k -3$$
Now: 


*

*$k = 1$: $3+2-2-3 \equiv_5 0$

*$k = 2$: $3\cdot 8 +2\cdot 4 -2\cdot2 -3 \equiv_5 -1 -2-4-3 \equiv_5 0$

*$k = 3$: $3\cdot 27 +2\cdot 9 -2\cdot3 -3 \equiv_5 6 +3-6-3 \equiv_5 0$
.......
