Prove by induction that $13n^{13} + 11n^{11}$ is divisible by $24$ for each $n \in \mathbb{N}$. I am currently tackling this question and I am stuck.
I have completed proven the base case where $P(1) = 13 + 11 = 24$ and $24\mid24$, hence base case is true.
Then I assume $P(k)$ to be true for some integer $k$, where $(24 \mid 13k^{13} + 11k^{11})$
Then I proceed on proving
$P(k+1) \mod 24$
$= 13(k+1)^{13}  + 11(k+1)^{11}  \mod 24$
$= (k+1)^{11} \cdot[13(k+1)^2 + 11]  \mod 24$
$= (k+11)^{11} \cdot [13k^2 + 26k + 24]   \mod 24$
Then I am very stuck here, as I don't know how do I remove $24$ from the equation; can I just remove $24$ from the equation since '24 will just mod 24 = 0' ?
Any help will be really appreciated.
Thank you.
 A: I know your question is to prove by induction and there is already an answer posted for that. So here is another way -
$13n^{13} + 11n^{11} = 13n^{13} - 13n^{11} + 13n^{11} + 11n^{11} = 13n^{11} (n-1)(n+1) + 24n^{11}$
So, we now just need to prove that $13n^{11} (n-1)(n+1)$ is divisible by 24.
If $n$ is even, either $(n-1), n$ or $(n+1)$ is divisible by $3$ and $n^{11}$ is divisible by $8$.
If $n$ is odd, one of $(n-1), n, (n+1)$ is again divisible by $3$. Also, both $(n-1)$ and $(n+1)$ are even and one of them is divisible by $4$.
So, it is evident that $13n^{13} + 11n^{11}$ is divisible by $24$ for all $n \in \mathbb{N}$
A: Claim:$\;$ $n^{m+2}-n^m$ is divisible by $24$ for $m\ge3$.
Here I prove that claim by induction, for $m$ odd, which is all that is needed for this problem.
Base case:  $n^5-n^3=n^3(n+1)(n-1)$ is divisible by $8$ and by $3$.
Induction step:  $n^{k+4}-n^{k+2}\equiv n^{k+2}n^2-n^{k+2}\equiv n^{k}n^2-n^{k+2}=0\bmod24.$
Therefore, $13n^{13}+11n^{11}\equiv13n^{11}+11n^{11}=24n^{11}\equiv0\bmod24$.
A: Put $\, m = 2^3\cdot 3,\ e = 11,\ j = 13,\ k = -11\,$ in the Euler-Fermat generalization below.
Theorem $\  $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{e_{1}}\cdots\:p_k^{e_k}\ $ and suppose that for all $\,i,\,$ $\ \color{#0a0}{e_i\le e}\ $ and $\ \phi(p_i^{e_{i}})\mid f\,$ and $\, j\equiv k\pmod{\!m}.$  Then $\ m\mid \color{#0a0}{a^e}(j\,a^f-k)\ $ for all $\: a\in \mathbb Z.$
Proof $\ $ Notice that if $\ p_i\mid a\ $ then $\:p_i^{e_{i}}\mid \color{#0a0}{a^e}\ $ by $\ \color{#0a0}{e_i \le e}.\: $ Else $\:a\:$ is coprime to $\: p_i\:$ so by Euler's phi theorem, $\!\bmod q = p_i^{e_{i}}:\, \ a^{\phi(q)}\equiv 1 \Rightarrow\ a^f\equiv 1\, $ by $\: \phi(q)\mid f\, $ and  modular order reduction, so $\,j\equiv k\,\Rightarrow\, j\,a^f-k\equiv 0.\,$ All prime powers $\, p_i^{e_{i}}\ |\  a^e (ja^f\! - k)\ $ so too does their lcm = product = $m$.
Examples $\ $ You can find many illuminating examples in prior questions, e.g. below
$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$
$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$
$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$
$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$
A: Another way is to express this polynomial in an alternative base, the binomial coefficients one: $$13n^{13}+11n^{11}=\sum\limits_{k=0}^{13} a_k\binom{n}{k}$$
Where $n^p = \sum\limits_{k=0}^{p} k!\, S(p,k)\,\binom{n}{k}\text{ and } S$ is the Stirling number of second kind.
And verify that all coefficients are divisible by $24$ :
$$a_k = k!\, \big(13\,S(13,k)+11\,S(11,k)\big)$$
Since $\ k!\ $ is divisible by $24$ as soon as $\ k\ge 4\ $, you only need to verify the first $4$ terms.

*

*$a_0 = 0$

*$a_1 = 24$

*$a_2 = 128976 = 24\times 5374$

*$a_3 = 22287816 = 24\times 928659$
Yet binomial coefficients being integers too,  the whole expression is divisible by $24$.
