Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5?

Question

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.

Answer 1

Alternative solution. Note that $$8n^3+12n^2-2n-3=(2n+3)(2n+1)(2n-1)$$ which is divisible by $5$ iff $(2n+3)$ or $(2n+1)$ or $(2n-1)$ is divisible by $5$. Now $$\begin{align} &2n+3\equiv 0\leftrightarrow 2n\equiv -3\equiv 2\leftrightarrow n\equiv 1 \pmod{5}\\ &2n+1\equiv 0\leftrightarrow 2n\equiv -1\equiv 4\leftrightarrow n\equiv 2 \pmod{5}\\ &2n-1\equiv 0\leftrightarrow 2n\equiv 1\equiv 6\leftrightarrow n\equiv 3 \pmod{5}\end{align}$$ Hence $8n^3+12n^2-2n-3$ is divisible by $5$ iff $n\equiv 1,2,3$ modulo $5$.

Answer 2

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:

• $k = 1$: $3+2-2-3 \equiv_5 0$
• $k = 2$: $3\cdot 8 +2\cdot 4 -2\cdot2 -3 \equiv_5 -1 -2-4-3 \equiv_5 0$
• $k = 3$: $3\cdot 27 +2\cdot 9 -2\cdot3 -3 \equiv_5 6 +3-6-3 \equiv_5 0$

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