Is this possible to prove through the induction method. It seems it is not to me. I built a base case, proceeded to substitute in k, then finally moved onto my $k+1$ case. Where I ended up with a polynomial that seems to be irreducible. My answer looks like
$Q(k+1) = 1+3+7+...+K+(K+1) = (K^3 + 11k + 6)/6$
From my understanding for this to be true all variables k on our right hand side must be simplified to (k+1) which I cannot seem to find a way to do.
You can prove it directly: notice that $$\begin{align}n^3 + 5n &\equiv n^3-n \pmod 6 \\&= (n-1)(n)(n+1) \pmod 6\end{align}$$
Now, at least one of $n-1,n,n+1$ must be divisible by 2. The same argument goes for 3. Hence both $2$ and $3$ must divide $n^3+5n$.
Hence $6= 2\cdot3$ must divide $n^3+5n$.
Proof:
Base case: $n=1$:
$((1)^3 + 5(1)) = 6$ which is divisible by 6.
Inductive step: Suppose that $(n^3 + 5n)$ is divisible by $6$, prove that $((n+1)^3 + 5(n+1))$ is divisible by 6.
Therefore: $((n+1)^3 + 5(n+1))$ = $(n^3 + 3n^2 + 3n + 1) +(5n + 5)$ $$=(n^3 + 5n) + 3(n^2 + n) + 6$$
Do you think you can take it from here?
clearly the statement is true for $k=1$ since, $k^3 +5k = 1 +5 = 6$
Assume that it is true for k implies $k^3 +5k$ is divisible by six.
and show that it is true for k+1
$$ \begin{align}(k+1) ^3 + 5(k+1) & = (k^3 + 3k^2 +3k +1) + (5k + 5) \\ & = (k^3 +5k) + (3k(k+1)) + 6 \end{align} $$
Now, the first term is divisible by six by the assumption in step 2. The middle term, $3k(k+1)$ is also divisible by six for all $k \ge 1$ (proof left to the reader) and clearly the last term is also divisible by six. So, the proposition is true for $k+1$. QED.
It can be done by induction: Let $f(n) = n^3 + 5n$, and suppose that it's divisible by $6$ for some $n$. Then
\begin{align*} f(n + 1) - f(n) &= (n + 1)^3 + 5(n + 1) - \Big(n^3 + 5n\Big) \\ &= 3n^2 + 3n + 1 + 5 \\ &= 3n(n + 1) + 6 \end{align*}
If you can convince yourself that this is divisible by $6$, you're done (why?).
You want to show that $n^3 + 5n \equiv 0 \pmod 6 $. As you already showed, the base case $n=1$ checks out. So let's look at $n=k+1$: $$ (k+1)^3 + 5 (k+1) \equiv 0 \pmod 6 $$ $$ k^3 + 3k^2 + 3k + 1 + 5k+5 \equiv 0 \pmod 6 $$ and we can get rid of the $1+5$ since it is divisible by $6$, $$ k^3 + 3k^2 + 8k \equiv 0 \pmod 6 $$
You can continuously alternate between factoring and expanding and you'll always be able to remove a factor of $6$.
If you are willing to check two base cases, then you can prove it by induction using
$$\begin{align} (n+2)^3+5(n+2)&=n^3+6n^2+12n+8+5n+10\\ &=n^3+5n+6(n^2+2n+3)\\ \end{align}$$
P = n(n^2 + 5) divisible by 2: If n is then P is, if n = 1 ( mod 2 ) then n^2 + 5 = 1 + 1 ( mod 2 ) = 0 ( mod 2 ).
P is divisible by 3. If n = 0 ( mod 3 ) , done. If not then either n = 1 ( mod 3 ), then n^2 + 5 = 1 + 2 ( mod 3 ) = 0 ( mod 3). If n = 2 ( mod 3 ) , then n^2 + 5 = 1 + 2 ( mod 3 ) = 0 ( mod 3 ). So P is divisible by 2 and 3 hence by 6.
Without induction or modular arithmetic,
Because $6 \mid (6n)$, we have that
$$ 6 \mid ((n^3 + 5n) - (n^3 - n)).$$
Now, $n-1$, $n$, and $n+1$ are consecutive integers so $6\mid(n-1)(n)(n+1)$. But
$$ (n-1)(n)(n+1) = n^3 - n.$$
So it must be that
$$ 6 \mid (n^3 -1).$$
We have that $6$ divides both $(n^3 + 5n) - (n^3 - n)$ and $n^3 - n$, so it also divides their sum:
$$ 6 \mid (((n^3 + 5n) - (n^3-n)) + (n^3-n)).$$
That is,
$$ 6 \mid (n^3 + 5n). $$
$n^3+5n=n^3-n+6n=(n+1)n(n-1)+6n$
It is obvious that $6\mid 6n$.
It is also obvious that $3! \mid (n+1)n(n-1) \implies 6\mid (n+1)n(n-1)$ as $(n+1),n,(n-1)$ are three consecutive integers.
