Prove that for all $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$.

I have shown the base case if $n=0$ (which is a natural number for my course).

I'm now trying to prove that $9k=4^n+6n-1$.

I substituted $n+1$ for $n$, and have $4^{(n+1)}+6(n+1)-1$, which equals $4*4^n+6n+5$.

I'm stuck. Is that not the right way to go? I don't know how to get from the “$+5$” to “$-1$” so I can substitute $9k$ to the right side of the equation. Any help would be greatly appreciated.

  • $\begingroup$ Do you know modular arithmetic, in particular arithmetic modulo $\,9\,$? $\ \ $ $\endgroup$ – Gone Dec 20 '13 at 18:38
  • $\begingroup$ @BillDubuque proving it like that he'll have 9 cases which is a bit long so I'd prefer a proof by induction which is more elegant. $\endgroup$ – Module Dec 20 '13 at 18:43
  • $\begingroup$ @Module Proofs of divisibility claims by modular arithmetic can be very efficient (and need not involve any case analysis). For example, that $\ 10\mid 98765 - 12345\ $ Is true simply because $\rm\ mod\ 10\!:\ 98765 - 12345\equiv 5 - 5\equiv 0,\,$ since $\rm\,10j+k\equiv k.\,$ $\endgroup$ – Gone Dec 20 '13 at 19:05
  • $\begingroup$ In his case the only way I see of approaching it is by cases. $\endgroup$ – Module Dec 20 '13 at 19:09
  • $\begingroup$ @Module Look further. For one example, take my proof and replace every occurrence of $\rm\, d\mid m\,$ by $\rm\, m\equiv 0\pmod d.\,$ Then the proof uses only arithmetic $\rm\, mod\ d,\,$ and is slightly clearer. I didn't do that since the OP did not confirm that they know modular arithmetic (congruences). $\endgroup$ – Gone Dec 20 '13 at 21:57

It's really simple this is how I'd approach it:


and since you've already proved that $4^n+6n-1=9k$ then you will have: $4(4^n+6n-1)-9(2n-1)=9k*4-9(2n-1)$

$=9(4k-(2n-1))$ and since $k$ and $2n-1$ are both integers then you can let $k'=4k-(2n-1)$ .

Now you'll have that $4^{n+1}+6(n+1)-1=9kk'$ which is a multiple of 9. So you can conclude for all natural numbers $n$ that $4^n+6n-1$ is divisible by $9$.

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  • $\begingroup$ Why $4^{n+1}+6(n+1)-1$ is equal to $9kk'$, shouldn't it be equal to $9k'$, since $k'$ is what we have just found that multiplies 9 and gives $4^{n+1}+6(n+1)-1$?? $\endgroup$ – nbro Nov 19 '14 at 23:23
  • $\begingroup$ @nbro Well no. We have already proved that $4^n+6n-1=9k$ in the induction step. So for the $n+1$ case you will get $9kk'$. $\endgroup$ – Module Nov 22 '14 at 9:06
  • $\begingroup$ I was not saying $4^n + 6n - 1 = 9k$, I meant that you should have $9k'$, why? Because you have found the $k' = 4k - (2n - 1)$ for $n + 1$... $\endgroup$ – nbro Nov 22 '14 at 21:07

hint $$4^{n+1}+6(n+1)-1=4\cdot4^n+6n+5=4(4^n+6n-1)-18n+9=$$ $$=4(4^n+6n-1)-9(2n-1)$$ first part is divisible to 9 by assumption and second part have 9 as factor

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Let $\displaystyle f(n)=4^n+6n-1$

So, $\displaystyle f(m+1)-f(m)=4^m(4-1)+6=3(4^m-1)+9$

Using Why $a^n - b^n$ is divisible by $a-b$? $4^m-1$ is divisible by $4-1=3$

So, $\displaystyle f(m+1)-f(m)$ is divisible by $9$

$\displaystyle\implies f(m+1)$ will be divisible by $9$ if $\displaystyle f(m)$ is

Now, establish the base case i.e., for $n=1$

As to the origin of the problem,

for $n\in\mathbb N$, $\displaystyle4^n=(1+3)^n=\sum_{0\le r\le n}\binom nr3^r=1+3n+\sum_{2\le r\le n}\binom nr3^r$

Observe that the last part is divisible by $3^2=9$ as Proof that a Combination is an integer

So, $\displaystyle4^n=1+3n+9k$(where $k$ is an integer $\ge0$ )

or in congruence form $\displaystyle4^n\equiv1+3n\pmod9\ \ \ \ (1)$

This can be derived using induction as follows

If $\displaystyle4^m=1+3\cdot\underbrace m+9\cdot\underbrace u$ (where $u$ is an integer),

$\displaystyle\implies4^{m+1}=4\cdot4^m=(1+3)(1+3m+9u)=1+3\underbrace{(m+1)}+9\underbrace{(u+3u+m)} $

Now $\displaystyle6n-1$ actually compensates the constant and the coefficient of $n$ in $(1)$

More generally, we can prove $\displaystyle(1+a)^n\equiv1+na\pmod{a^2}$

In that case, the compensator will be $\displaystyle(a^2-a)n-1\pmod{a^2}$

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Below we prove a more general result, because it requires little extra work yet yields great rewards, revealing structure that allows us to see the Binomial Theorem as the essence of the matter.

$\!\!\begin{align}\rm{\bf Theorem}\ \ \ \ \forall n\in\Bbb N\!:\ d\mid f(n) = a^n\! + bn + c &\rm \iff d\mid \color{blue}{(a\!-\!1)^2},\, \color{brown}{a\!+\!b\!-\!1},\, \color{darkorange}{1\!+\!c}\\ &\rm \iff d\mid f(0),\,f(1),\,f(2)\end{align}$

Proof $\ (\Leftarrow)\ $ By induction on $\rm\,n.\,$ The base case $\rm\,n=0\,$ claims that $\rm\, d\mid f(0)= \color{darkorange}{1\!+\!c},\,$ which is true by hypothesis. $ $ Next, assume that $\rm\,\color{#c00}{d\mid f(n)}\,$ as our inductive hypothesis.

Note $\rm\ \ \color{#0a0}{d\mid f(n\!+\!1)-f(n)}\, =\, a^n(a-1) + b\, =\, \color{blue}{(a^n\!-\!1)(a\!-\!1)}\, +\, \color{brown}{a\!+\!b\!-\!1}\ $ by hypotheses.

Thus $\rm\ \ \color{#0a0}{d\mid f(n\!+\!1)-f(n)}\ $ and $\rm\ \color{#c00}{d\mid f(n)}\ $ so $\rm\ d\,$ divides their sum $\rm\,f(n\!+\!1).$

$(\Rightarrow)\,\ \ \rm\, f(0)= \color{darkorange}{1\!+\!c},\,$ $\rm\ f(1)\!-\!f(0) = \color{brown}{a\!+\!b\!-\!1},\ $ $\rm f(2)\!-\!f(1)-(f(1)\!-\!f(0)) = (\color{blue}{a\!-\!1})^2\ \ $ QED

Specializing $\rm\ a,b,c,d = 4,6,-1,9\,$ yields the original problem:

$\rm\qquad\qquad\ 9\mid \color{blue}{(4\!-\!1)^2},\, \color{brown}{4\!+\!6\!-\!1},\,\color{darkorange}{1\!-\!1}\ \ \Rightarrow\ \ 9\mid f(n) = \color{blue}4^n\! + \color{brown}6n \color{darkorange}{-1},\ \ for\ all\,\ n\in\Bbb N$

The relationship with the Binomial Theorem ($\rm\color{#0a0}{BT}$) becomes obvious if we rewrite the theorem in arithmetical (equational) language, by replacing divisibility relations by equivalent congruences. Rearranging $\rm\ f(n) = a^n\! + b\,n + c\equiv 0\pmod d\, $ we have

$$\begin{eqnarray}\rm mod\ d\!:\ a^n \equiv \color{darkorange}{-c}+n(\color{brown}{-b}) &\equiv&\rm \color{darkorange}1 + n(\color{brown}{a\!-\!1})\quad by\ \ \color{brown}{{-}b\equiv a\!-\!1},\ \ \color{darkorange}{c\equiv -1}\\ \rm but\ \ \ \ a^n\equiv\, (1+a\!-\!1)^n&\rm\overset{\color{#0a0}{BT}}\equiv&\rm 1 + n(a\!-\!1) + \color{blue}{(a\!-\!1)^2}(\cdots)\ \ and\ \ \color{blue}{(a\!-\!1)^2}\equiv 0 \end{eqnarray}$$

Thus the result my be discovered by binomial expansion, then truncating higher powers of $\rm\,a\!-\!1.\,$ Hence the induction in the above proof is essentially a special case of the type of induction that is used to prove the first two terms of the Binomial Theorem. Indeed, one could rewrite the above proof to make this relationship more explicit. I leave that as an instructive exercise for the reader.

See also here for the slight generalization $\rm\, f(n) = e \,a^n+\cdots\,$ vs. $\,a^n+\cdots$

Generally as explained here, we deduce that $\rm\,f_n = f(n)\,$ satisfies a monic order $3$ recurrence $\,\rm (S-1)^2(S-a)\,f_n = 0\ $ so $\rm\ f_{n+3} = c_2\, f_{n+2} + c_1\, f_{n+1} + c_0\, f_n\,$ for $\rm \,c_i\in\Bbb Z,\,$ so a simple induction shows $\rm\,f_0,f_1,f_2\equiv 0\iff f_n\equiv 0\,\ \forall n\in \Bbb N,\,$ so the above proof may be viewed as a special case of the uniqueness theorem for recurrences. Thus $\ \rm d\mid f_0,f_1,f_2,\,\ldots\!\iff d\mid f_0,f_1,f_2,\,$ so $\,\rm \gcd(f_0,f_1,f_2,\ldots) = \gcd(f_0,f_1,f_2)$

Those who know a little ring theory may find it useful to view the universal essence of the matter as nothing but the Binomial Theorem in the ring of dual numbers $\rm\,R[x]/x^2.\,$ This ring proves handy when algebraically studying (higher) derivations, tangent/jet spaces, etc.

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  • 4
    $\begingroup$ Tis true, I am not the one to judge this. But how on earth this grand mixture of colors is helpful, and not anything but an eyesore? (Personally, that is the definition of going too far.) $\endgroup$ – Asaf Karagila Dec 20 '13 at 22:45
  • $\begingroup$ @Color-haters It is quite sad that some users downvote answers apparently because they don't like use of color. From many years of feedback it is clear that color has helped many readers more easily follow the logical flow of arguments. $\endgroup$ – Gone Sep 12 '19 at 14:09

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