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I'll state the question here:

If "$P(n): 2 \cdot 4^{2n + 1} + 3^{3n + 1}$ is divisible by $\lambda$, for all $n \in N$" is true, then what is the value of $\lambda$?

This is what I can think of:

$P(1): 2 \cdot 4^{2(1) + 1} + 3^{3(1) + 1}$ is divisible by $\lambda$.

i.e., $209$ is divisible by $\lambda$.

And,

$P(2): 2 \cdot 4^{2(2) + 1} + 3^{3(2) + 1}$ is divisible by $\lambda$.

i.e., $4235$ is divisible by $\lambda$.

Now the common factors of the $209$ and $4235$ are $1$ and $11$. So, $\lambda$ is one of $1$ and $11$. My textbook says that the answer is 11. And this how it was solved in the book.

Now I can't understand why can't the answer be $1$? Is doing what I did above enough to state the answer as $11$? What if for some $k \in N$, $2 \cdot 4^{2n + 1} + 3^{3n + 1}$ is not divisible by $11$? So to avoid missing such a term, I believe I should substitute $\lambda$ with $11$ in the statement and prove the statement using induction (that is what the chapter in my book is about). If I am able to prove it (and I will be since it's true), the answer is $11$. Otherwise, the answer is $1$.

My doubt is regarding the approach to the question. Does the method by which I solved it above suffice to give the answer as $11$? Or, as I mentioned in the previous paragraph, do I have to prove the statement again by replacing $\lambda$ with $11$?

Shinsekai no Kami
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  • Most powers of $2.4$ are not integers ... – Hagen von Eitzen Feb 05 '17 at 15:27
  • I assume you mean to write $A_n=2\times 4^{2n+1}+3^{3n+1}$, and you seek to show that $11,|,A_n,\forall n$, yes? If so, then you just need to check $A_n$ for $n\in {0,\cdots,9}$ . It helps to note that both $3,4$ have order $5\pmod {11}$. – lulu Feb 05 '17 at 15:44
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    I hope I didn't do anything wrong by editing .. –  Feb 05 '17 at 15:44
  • @HagenvonEitzen I meant $2 \cdot 4$. Sorry for the formatting error. I wasn't aware. – Shinsekai no Kami Feb 07 '17 at 14:38
  • @lulu I didn't get your point. Probably cause I'm not aware of modular arithmetic. Please check out the question again. I edited it and made it clear what I really want to ask. – Shinsekai no Kami Feb 07 '17 at 14:41
  • Well, perhaps I do not understand. I thought that you (sensibly) checked a few cases out and determined empirically that the answer had to be $\lambda= 11$. Great! Now you have to prove it. Are you saying that you don't know how to work $\pmod {11}$? If so...well, there's probably a trick that avoids that. Let me think a bit. – lulu Feb 07 '17 at 15:06
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    Looking at the posted solutions, the answer from @BillDubuque seems to dodge modular arithmetic effectively. – lulu Feb 07 '17 at 15:11
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    As the question is written, there is nothing to exclude $\lambda=1$. Clearly they intended you to find the maximum value of $\lambda$ and the answers show how. If I got this question I would be tempted to answer $\lambda=1$ will work because $1$ divides all naturals and quit at that point. Of course, if my teacher did not have a sense of humor I might get rather few points. – Ross Millikan Feb 09 '17 at 18:20

4 Answers4

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You've proved $\lambda$ divides $11$. To see that $11$ divides $\lambda$, we need to show that $11$ divides $f(n)$ for all $n$. To see this let's use induction on $n$, and note that $f(n) = g(n) + h(n)$ where: $$g(0) = 8\\ h(0) = 3\\ g(n+1) = 16g(n) = 11g(n)+ 5g(n)\\ h(n+1) = 27h(n) = 11\cdot 2h(n) + 5h(n) $$ So $11$ certainly divides $f(0) = 11$ and we have $$ f(n+1) = 11(g(n) + 2h(n)) + 5(g(n) + h(n)) = 11(g(n)+2h(n)) + 5f(n). $$

Hence, if $11$ divides $f(n)$, it will also divide $f(n+1)$. So $11$ divides $f(n)$ for all $n$ by induction.

Rob Arthan
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  • Thanks! I couldn't understand the last sentence. Probably cause I'm not aware of modular arithmetic. Please check out the question again. I edited it and made it clear what I really want to ask. – Shinsekai no Kami Feb 07 '17 at 14:45
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    Just showing that $11$ divides the first two values of $f(n)$ doesn't obviously guaranteee that if divides $f(n)$ for every $n$, so the proof you quote from your book is incomplete. I've tried to free my answer of modulo arithmetic.. – Rob Arthan Feb 07 '17 at 17:51
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Hint $\ {\rm mod}\ 11\!:\,\ 8\cdot 16^n + 3\cdot \overbrace{ \color{#c00}{27}^n}^{\textstyle\color{#c00}{16}^n}\equiv 11\cdot 16^n\equiv 0\ $ by $\ \color{#c00}{27\equiv 16}\,$ & Congruence Power Rule.

Remark $ $ If modular arithmetic is unfamiliar then we can instead rewrite is as follows

$$ 3(27^n\!-16^n) + 11\cdot 16^n$$

which is divisible by $11$ by $\,11 = 27\!-\!16\mid 27^n\!-16^n$ (provable by induction or by invoking the Factor Theorem $\, a-b\mid a^n-b^n)$

Induction is also (implicitly) used in the modular proof because it uses $\,27\equiv 16\,\Rightarrow\, 27^n\equiv 16^n\,$ and the proof of this Congruence Power Rule is by induction on $\,n$.

Update $ $ OP seeks a direct inductive proof without modular arithmetic etc. Below is one.

$$\begin{align} f(n)\ &=\qquad 8\cdot 16^n +\, 3\cdot 27^n\\ \Rightarrow\ f(n\!+\!1)\ &=\, 8\cdot \color{#0a0}{16}\cdot 16^n +\, 3(\color{#0a0}{16}\!+\!11)\, 27^n\\ &=\, \color{#0a0}{16}\ \color{#c00}{f(n)}\ \ \ +\ \ \ 3\cdot \color{}{11}\cdot 27^n\\ &=\, \color{#c00}{11} (\color{#0a0}{16}\, \color{#c00}m\ \ +\,\ \ 3\cdot 27^n)\ \ {\rm by}\,\ \color{#c00}{f(n) = 11m}\,\ {\rm(inductive\ hypothesis)} \end{align}$$

therefore $\ f(n) = 11m\,\Rightarrow\, f(n\!+\!1) = \color{#c00}{11}k,\ $ i.e. $\,11\mid f(n)\,\Rightarrow\, 11\mid f(n\!+\!1)$.

Bill Dubuque
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  • Thanks! Please check out the question again. I edited it and made it clear what I really want to ask. – Shinsekai no Kami Feb 07 '17 at 14:47
  • @Sam By above we know $d=11$ divides all the $a(i)$. By your argument we know that $d$ is the greatest common divisor of the $a(i)$ since $d\mid a(1),a(2),\Rightarrow,d\mid\gcd(a(1),a(2)) = 11.\ $ – Bill Dubuque Feb 07 '17 at 15:32
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    @Sam It seems the root of your question is that, as written, the question doesn't ask for the greatest common divisor (gcd) so both $1$ and $11$ are common divisors. Usually such questions ask for the gcd (the other common divisors are precisely the divisors of the gcd). – Bill Dubuque Feb 07 '17 at 16:46
  • Yes. But I can't be sure if $\lambda = 11$ for all natural numbers just because it is the case for $n = 1 , 2$. So, I have to use induction or something like you did. Well, thanks again! – Shinsekai no Kami Feb 09 '17 at 14:12
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Note that any numbers of the form $u_n=Aa^n+Bb^n$ satisfy the recurrence $u_{n+1}=(a+b)u_n-abu_{n-1}$ (easy to check directly). You can use this to show that any number which is a factor of $u_1$ and $u_2$ is a factor of $u_n$ for $n\ge 1$.

Here you use $a=16, b=27$

This works as follows. $a,b$ are the roots of the equation $(x-a)(x-b)=0$ so that $x^2=(a+b)x-ab$. Multiply through by $x^{n-1}$ and this gives $x^{n+1}=(a+b)x^n-(ab)x^{n-1}$. This still has roots $a$ and $b$ so setting $x=a$ and multiplying through by $A$ gives $$Aa^{n+1}=A(a+b)a^n-A(ab)a^{n-1}$$ Similarly with $b, B$$$Bb^{n+1}=B(a+b)b^n-B(ab)b^{n-1}$$

Adding these together gives $$Aa^{n+1}+Bb^{n+1}=(a+b)(Aa^n+Bb^n)-(ab)(Aa^{n-1}+Bb^{n-1})$$ which is just $$u_{n+1}=(a+b)u_n-abu_{n-1}$$

Mark Bennet
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  • Thanks! But it's a bit inconvenient for me to remember this for examinations since I've never used it before. And I don't know how to prove it with this either, though I got your point. Please check out the question again. I edited it and made it clear what I really want to ask. – Shinsekai no Kami Feb 07 '17 at 14:53
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    @SamInuyashaANMF The proof is easy by induction - once you have two consecutive terms divisible by the same number, divisibility inevitably persists You can remember the form easily from $(x-a)(x-b)=x^2-(a+b)x+ab$. But you probably don't want that overload if it is new to you and you are in exam prep mode. – Mark Bennet Feb 07 '17 at 15:02
  • I don't get how are the two related. That's $x^2$, $x$ and $ab$ and that's $u_{n+1}$, $u_n$ and $u_{n-1}$. – Shinsekai no Kami Feb 07 '17 at 15:08
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    @SamInuyashaANMF I've done a little note at the end of the answer to show how these things are related. – Mark Bennet Feb 07 '17 at 15:53
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Take it modulo 11:

$2\cdot4^{2n + 1} + 3^{3n + 1} = 8 \cdot 5^n + 3 \cdot 5^n = 11 \cdot 5^n = 0$

  • Thanks! But sorry, I'm not aware of modular arithmetic. Please check out the question again. I edited it and made it clear what I really want to ask. – Shinsekai no Kami Feb 07 '17 at 14:54