Prove by induction that $13n^{13} + 11n^{11}$ is divisible by $24$ for each $n \in \mathbb{N}$.

Question

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.

Answer 1

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 $n \in \mathbb{N}$

Answer 2

Nice effort, your work got most of the way home. Using induction, and correcting a typo in the last (modular) equation in your query, you (effectively) reduced the problem to showing that

$[E_1]\; (k + 1)^{(11)} \times [13(k^2 + 2k)] \;\equiv\; 0 \pmod{24}.$

This can be done by showing that the product of the two factors in $E_1$ must be both a multiple of 8 and a multiple of 3.

$\underline{\text{multiple of 3}}$

The residue mod 3 of $k$ will be either 0, 1, or 2.

The residue of 2 is immediately disposed of by noting the first factor in $E_1, (k+1)^{(11)}.$

The residue of 0 is immediately disposed of by noting that $k$ divides the 2nd factor in $E_1, 13(k^2 + 2k).$

The residue of 1 is disposed of by noting that
if $k \equiv 1 \pmod{3},$
then $k^2 \equiv 1 \pmod{3},$ and
$2k \equiv 2 \pmod{3}.$

Thus, $k \equiv 1 \pmod{3} \Rightarrow (k^2 + 2k) \equiv 0 \pmod{3}.$

$\underline{\text{multiple of 8}}$

$k$ odd is disposed of by noting the first factor in $E_1, (k+1)^{(11)}.$

If $k \equiv 0 \pmod{4},$
then $8|k^2$ and $8|2k \Rightarrow 8|(k^2 + 2k).$

The only remaining case is $k \equiv 2 \pmod{4}.$
Noting that $(4r + 2)^2 = (16r^2 + 16r + 4),$ this means that
$k \equiv 2 \pmod{4} \Rightarrow k^2 \equiv 4 \pmod{8}.$

Further, $k \equiv 2 \pmod{4} \Rightarrow 2k \equiv 4 \pmod{8}.$

Thus, $k \equiv 2 \pmod{4} \Rightarrow $
$(k^2 + 2k) \equiv (4 + 4) \pmod{8}.$

Addendum Reaction to stochasticboy321's subsequent comment.
Proof by induction.

The short answer is : see J.W.Tanner's answer, which I consider a much superior induction approach than my induction approach (below). Rather than delete the remainder of this addendum, I'll leave it in place for contrast.

Lemma 1
Given $k, r, n, a \,\in \mathbb{Z^+},$ with $r \equiv 0 \pmod{a},$
then $(k + r)^n \equiv k^n \pmod{a}.$

Proof
In the binomial expansion of $(k + r)^n - k^n,$
$r$ divides every term.

Lemma 2
For $k \in \{1, 2, 3\}, k^{(13)} - k^{(11)} \equiv 0 \pmod{3}.$

Proof
By inspection, or you can notice that
$(k^{(13)} - k^{(11)}) = k^{(11)}(k-1)(k+1).$

Lemma 3
For $k \in \{1, 2, 3, 4, 5, 6, 7, 8\}, k^{(13)} - k^{(11)} \equiv 0 \pmod{8}.$

Proof
By inspection, or you can (again) notice that
$(k^{(13)} - k^{(11)}) = k^{(11)}(k-1)(k+1).$

Note
I definitely regard the concept in Teresa Lisbon's comment as the natural way to craft a proof by induction. Unfortunately, I could find no way of doing this that did not descend into the ugliness of both $(k+1)^{(11)}$ and $(k+1)^{(13)}.$ Given how much easier a non-induction proof is, I am forced to regard the raw descent into these binomial expansions as out of bounds.

I admit that my use of Lemma's 2 and 3 above to craft a proof by induction is highly artificial, since the analysis in these lemmas lends itself so easily to a non-inductive proof. However, this is the only reasonable way that I could conjure a proof by induction.

Let $f(n)$ denote $13n^{(13)} + 11n^{(11)} : n \in \mathbb{Z^+}.$

Claim-1 $\forall n \in \mathbb{Z^+}, f(n) \equiv 0 \pmod{3}.$

Proof
Since $13 \equiv 1 \pmod{3},$ and $11 \equiv (-1) \pmod{3},$
$f(n) \equiv n^{(13)} - n^{(11)} \pmod{3}.$
Therefore, Lemma 2 establishes that Claim-1 holds in the
base cases of $n \in \{1, 2, 3\}.$
Then, Lemma 1 may be invoked to establish, by induction,
that Claim-1 holds for all $n \in \mathbb{Z^+}.$

Claim-2 $\forall n \in \mathbb{Z^+}, f(n) \equiv 0 \pmod{8}.$

Proof
Since $13 \equiv 5 \pmod{8},$ and $11 \equiv (-5) \pmod{8},$
$f(n) \equiv (5) \times \left[n^{(13)} - n^{(11)}\right] \pmod{8}.$
Therefore, Lemma 3 establishes that Claim-2 holds in the
base cases of $n \in \{1, 2, 3, 4, 5, 6, 7, 8\}.$
Then, Lemma 1 may be invoked to establish, by induction,
that Claim-2 holds for all $n \in \mathbb{Z^+}.$

Answer 3

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, $13n^{13}+11n^{11}\equiv13n^{11}+11n^{11}=24n^{11}\equiv0\bmod24$.

Answer 4

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

$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$

$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$

$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$

$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$

Answer 5

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 $24$.