Prove that $9\mid (4^n+15n-1)$ for all $n\in\mathbb N$

Question

First of all I would like to thank you for all the help you've given me so far.

Once again, I'm having some issues with a typical exam problem about divisibility. The problem says that:

Prove that $\forall n \in \mathbb{N}, \ 9\mid4^n + 15n -1$

I've tried using induction, but that didn’t work. I've tried saying that:

$4^n + 15n-1 \equiv 0 \pmod{9}$. Therefore, I want to prove that $4^{n+1} + 15(n+1) -1 \equiv 0 \pmod{9}$.

I've prooved for $n=1$, it's $18\equiv 0 \pmod{9}$, which is OK.

But for the inductive step, I get:

$4\cdot4^n + 15n+15-1 \equiv 0 \pmod{9}$

And from there, I don't know where to replace my inductive hypothesis, and therefore, that's why I think induction is not the correct tool to use here. I guess I might use some tools of congruence or divisibility, but I'm not sure which ones.

I do realize that all $n\in \mathbb{N}/ \ 3 \ |\ n \Rightarrow 4^n \equiv 1 \pmod{9} \text{ and } 15n \equiv 0 \pmod{9}$. In that case, where 3 divides n, then I have prove that $4^n + 15n-1 \equiv 0 \pmod{9}$. But I don't know what to do with other natural numbers that are not divisible by 4, that is, all $n \in \mathbb{N} / n \equiv 1 \pmod{3} \text{ or } n \equiv 2 \pmod{3}$.

Any ideas? Thanks in advance!

Find the remainder of $4^{3k+1}$ modulo $9$, and that of $4^{3k+2}$. Since you already know $4^{3k} = 1 + 9\cdot x$, that's not difficult. — Daniel Fischer, Jul 15 '13 at 14:47
You can do a proof by induction here you just need to show the additional result that $4^k+5 = 0$ mod $3$. — Cameron Williams, Jul 15 '13 at 14:48

score 10 · Accepted Answer · answered Jul 15 '13 at 14:50

10

By the Inductive Hypothesis, $4^n + 15n -1 \equiv 0$ so $4^n \equiv 1-15n$ and thus $$ 4^{n+1}+15(n+1)-1 = 4 \cdot 4^n + 15n + 14 \equiv 4 \cdot (1-15n) + 15n + 14 = 18 -45n \equiv 0 $$ since both $18$ and $45$ are divisible by 9.

answered Jul 15 '13 at 14:50

gt6989b

54,422

Thanks! I liked this answer by induction... – pmartelletti Jul 15 '13 at 14:53

score 6 · Answer 2 · answered Jul 15 '13 at 14:50

6

You can use the fact that any natural number can be written in three forms : $n = 3k$ or $n = 3k+1$ or $n = 3k+2$.

In the case : $n = 3k$

$$ 4^n + 15 n - 1 = 64^k + 45 k -1 \equiv 0 \pmod{9} $$

And then you do the same for the other cases

answered Jul 15 '13 at 14:50

Ayoub Ennassiri

454

André Nicolas · Answer 3 · 2013-07-15T15:02:47.867

3

Consider $$4\left(4^n+15n -1\right)-\left(4^{n+1} +15(n+1)-1\right) .$$

edited Jul 15 '13 at 15:02

answered Jul 15 '13 at 14:56

André Nicolas

507,029

score 3 · Answer 4 · edited Apr 13 '17 at 12:20

3

We don't really need induction as $$4^n+15n-1=(1+3)^n+15n-1$$ $$=1+\binom n13+\binom n23^2+\cdots+\binom n{n-1}3^{n-1}+3^n+15n-1$$ using Binomial Expansion

$$\implies 4^n+15n-1=18n+3^2\left(\binom n2+\cdots+\binom n{n-1}3^{n-3}+3^{n-2}\right)$$ which is clearly multiple of $9$ as we know Binomial coefficients are integers for positive integer $n$

edited Apr 13 '17 at 12:20

Community

1

answered Jul 22 '13 at 18:45

lab bhattacharjee

274,582

score 1 · Answer 5 · answered Jul 15 '13 at 14:52

1

You can do it by induction, skipping three steps. First check that it is correct for $n=1,2,3$. Then $4^{n+3}+15(n+1)-1=4^3\cdot 4^n+15n-45-1.$ As you have shown $4^3 \equiv 1 \pmod 9$ you are home.

answered Jul 15 '13 at 14:52

Ross Millikan

374,822

Below is a rigorous proof of the above claim. Dividing $,n,$ by $,3,$ we have $$\large \begin{align}n=r!+!3q,\Rightarrow,\bmod 9!:\ 4^n+15n &= 4^{r+3q} + 15(r!+!3q)\ &\equiv 4^r (\color{#c00}{4^3})^q + 15r + \color{#0a0}{45}q,\equiv, 4^r+15r\ &\ \ {\rm by}\ \ \ \ \color{#c00}{4^3\equiv 1},\ \ {\rm and}\ \ \color{#0a0}{45\equiv 0}\end{align}\qquad$$
where we used the Congruence Sum and Product Rules.
– Bill Dubuque Nov 08 '21 at 02:40
Alternatively $\bmod 9!:\ 4^{\large\color{#c00}3}!\equiv 1\Rightarrow 4^{\large n}\equiv 4^{:!\large n\bmod \color{#c00}3}!\equiv 4^{:!\large r},$ by MOR = mod order reduction. The proof in the prior comment essentially repeats inline the proof of MOR in this special case. – Bill Dubuque Nov 09 '21 at 20:15

score 1 · Answer 6 · answered Jul 15 '13 at 15:29

1

Note that $4^3\equiv1\pmod{9}$ and $15\cdot3\equiv0\pmod{9}$

Therefore, we only need to verify $4^n+15n\equiv1\pmod{9}$ for $n\in\{0,1,2\}$ since any $n$ is equivalent to one of these $\bmod{\,3}$. Since $\{1,19,46\}$ are all $\equiv1\pmod{9}$, the equation holds for all $n$.

answered Jul 15 '13 at 15:29

robjohn

345,667

This is a dupe of Ross's answer $37$ minutes prior. See my comment there for a rigorous proof of the above claim. – Bill Dubuque Nov 08 '21 at 02:39

score 0 · Answer 7 · answered Jul 15 '13 at 14:57

I'm going to do it by induction. First I need to prove that $4^k+5=0$ mod $3$.

Base case: $4^1+5 = 9$ and clearly $3$ divides this quantity.

Inductive hypothesis: $3|4^k+5$. Then $4^{k+1}+5 = 4*4^k+5 = 4^k+5+3*4^k$. Since $3|4^k+5$, it divides $4^{k+1}+5$ and we have shown the result.

Now we wish to prove the overall result that $3|4^k+15k-1$. Clearly the base case is true so we'll work on the inductive step. Suppose $3|4^k+15k-1$. Then $4^{k+1}+15(k+1)-1 = 4*4^k+15k+15-1 = (4^k+15k-1)+3(4^k+5)$. Since $3$ divides both of these terms individually, it divides their sum and we are done.

score -2 · Answer 8 · edited Sep 09 '14 at 11:58

-2

Consider $$4^n=9q-15n+1;$$ So $$4.4^n+15n+15-1=4(9q-15n+1)+15n+14$$ And go on.

edited Sep 09 '14 at 11:58

user37238

4,017

answered Sep 09 '14 at 11:37

Nathan

1

Prove that $9\mid (4^n+15n-1)$ for all $n\in\mathbb N$

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