$\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$

Question

Show $\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$

Indeed,

we've to show $5n^3+n=0\pmod 6 $

note that for $n\in \{0,1,2,3,4,5\}$ we've: $n^{3}=n\pmod 6$ then

$5n^{3}=5n\pmod 6 \implies 5n^{3}+n=6n\pmod 6=0\pmod 6$

AM i right ?

Answer 1

Yes, you're right. ${}{}{}{}{}{}{}{}{}{}{}{}{}$

Answer 2

Hint

Prove that $5n^3+n\equiv0\mod 2$ and $5n^3+n\equiv0\mod 3$ and notice that $2$ and $3$ are coprime.

Answer 3

Here's another way to do it: note that $5n^3+n = n(5n^2+1)$. $2$ divides this since one of $n$ and $5n^2+1$ is even.

If $n = 3k$, then we're done since $3$ will divide $n(5n^2+1)$.

If $n=3k+1$, then $5n^2+1 = 5(3k+1)^2+1 = 3[5(3k^2+2k)+2]$. So $3$ divides $n(5n^2+1)$. The remaining case is similar to check.

Answer 4

Use induction

$$ 5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 \Big( \tfrac{5}{2} n(n+1) +1 \Big) $$

and as $n(n+1)$ is even we see that

$$ 5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 k $$

As it is true for $n=1$, it is true for all $n \ge 1$.

Answer 5

Yes, it's correct. Alternatively $\,5(n^3\!-n)+6n\, =\, 5\underbrace{(n\!+\!1)n(n\!-\!1)}_{\large m} + 6n\, =\, 5\cdot \color{#c00}6{\large {n\!+\!1\choose 3}}\!+\color{#c00}6n$

Or, since $\,m\,$ is a product of $3$ consecutive integers, $\,2,3\mid m\,\Rightarrow\, 6={\rm lcm}(2,3)\mid m$