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I can see the answer to this in my textbook; however, I am not quite sure how to solve this for myself . . . the book has the following:

To take advantage of the inductive hypothesis, we use these steps:

$ 7^{(k+1)+2} + 8^{2(k+1)+1} = 7^{k+3} + 8^{2k+3} $

$$ = 7\cdot7^{k+2} + 8^{2}\cdot8^{2k+1}\\ = 7\cdot7^{k+2} + 64\cdot8^{2k+1}\\ = 7(7^{k+2}+8^{2k+1})+57\cdot8^{2k+1}\\ $$

While the answer is apparent to me now; how exactly would I go about figuring out a similar algebraic manipulation if I were to see something like this on a test? Is there an algorithm or a way of thinking about how to break this down that I'm missing? I think I'm most lost regarding the move from the second to last and last equations.

Source: Discrete Mathematics and its Applications (7th ed), Kenneth H. Rosen (p.322)

Anton Rasmussen
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    Also, you could observe that $8 = 1 + 7$ and $8^2 \equiv 7 \pmod {57}$, which means that the expression becomes $7^{n+2} + 8^{2n+1} \equiv 7^{n+2} + 7^n(1 + 7) \equiv 7^n(7^2 + 7 + 1) \equiv 7^n * 57 \equiv 0 \pmod {57}$. – Tob Ernack Jul 14 '17 at 17:14
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    Your aim is to somehow use your inductive hypothesis to deduce something about the $k+1$th case. ie. you're aiming to rearrange $7^{(k+1)+2} + 8^{2(k+1)+1}$ to include a $7^{k+2} + 8^{2k+1}$ term in it. If you do that, you'll see that you have to follow similar steps the book does. – Shuri2060 Jul 14 '17 at 17:14
  • See https://math.stackexchange.com/questions/2355529/induction-divisibility/2355774#2355774 – lab bhattacharjee Jul 14 '17 at 17:45

8 Answers8

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The key is to get very clear on what you have, and what you want.

For the inductive step, you have the inductive assumption:

$7^{k+2}+8^{2k+1}$ is divisible by $57$

And you want to show that:

$7^{(k+1)+2}+8^{2(k+1)+1}$ is divisible by $57$

So, you should think to yourself: OK, I want to show that $7^{(k+1)+2}+8^{2(k+1)+1}$ is divisible by $57$, and of course at some point I want to use my inductive hypothesis that $7^{k+2}+8^{2k+1}$ is divisible by $57$ to help me with this. So: I probably want to manipulate/rewrite the expression $7^{(k+1)+2}+8^{2(k+1)+1}$ so that I get some term $7^{k+2}+8^{2k+1}$ in there. How can I do that? Well, I have $1$ extra $7$ in the $7^{(k+1)+2}$ term, and I have $2$ extra $8$'s in the $8^{2k+1}$ term, so I should pull those out:

$$7^{(k+1)+2}+8^{2(k+1)+1} =$$

$$7^{(k+2)+1}+8^{(2k+1)+2} =$$

$$7\cdot 7^{k+2}+8^2\cdot 8^{2k+1}=$$

$$7\cdot 7^{k+2}+64\cdot 8^{2k+1}$$

Now, it is not quite clear yet how to isolate a $7^{k+2}+8^{2k+1}$, but at this point you might notice something: we got a 7 ... and we got a 64 ... the difference of which is 57 (!). Hmmm.... OK, let's try this:

$$7\cdot 7^{k+2}+64\cdot 8^{2k+1}=$$

$$7\cdot 7^{k+2}+(57+7)\cdot 8^{2k+1}=$$

$$7\cdot 7^{k+2}+57\cdot 8^{2k+1}+7\cdot 8^{2k+1}=$$

(and at this point you'll see it ...)

$$57\cdot 8^{2k+1}+7\cdot 7^{k+2}+ 7\cdot 8^{2k+1}=$$

$$57\cdot 8^{2k+1}+7\cdot (7^{k+2}+ 8^{2k+1})$$

The term $57\cdot 8^{2k+1}$ is clearly divisible by $57$, and we just isolated the $7^{k+2}+ 8^{2k+1}$ for the second term, so that is divisble by $57$ as well, meaning that their sum is divisible by $57$ as well.

User1331
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Bram28
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    Best answer imho. Really shows a possible thinking patern! – Student Jul 14 '17 at 17:19
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    @Student I appreciate that ... and it is how I thought about it myself. I was initially confused on that last step in the OP's posted solution as well ... and so maybe the whole problem was not with the induction part of the proof, but really that the posted solution took a step that was just a little too big for comfort for many people. – Bram28 Jul 14 '17 at 17:24
  • @Bram28 ... This is great! Thank you!! I should have waited longer to accept the answer; didn't think I'd get this much response!! – Anton Rasmussen Jul 14 '17 at 17:28
  • @AntonRasmussen Glad it helped! :) – Bram28 Jul 14 '17 at 17:30
  • @Student Re: "thinking pattern", if you sincerely desire to understand the (arithmetical essense) of the "pattern" in the induction, then great insight is gained by using congruence language, e.g. see my answer where I show that the induction on $n$ boils down to raising a congruence to the $n$'th power, allowing us to understand the whole induction in a single glance. This viewpoint makes such inductive proofs very simple. – Bill Dubuque Jul 14 '17 at 19:44
  • @BillDubuque i agree, but I meant it in view of the specific question for an induction proof as requested by the OP. Edit: did not see that your answer also uses induction. I like your answer too, but this one really gives the idea of a thinking process to me (probably also due to the way it is written) – Student Jul 14 '17 at 19:49
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    @Student If you try to solve more complicated problems like this then you will soon be lost if you do not learn to exploit the innate arithmetical structure (here the inductive extension of the Congruence Product Rule, i.e. the Congruence Power Rule). – Bill Dubuque Jul 14 '17 at 20:02
  • @BillDubuque totally agree on that, I really do! But in view of the question asked (with the provided solution from the notes) I think this answer is great. Still think your right on the congruences anyway :) – Student Jul 14 '17 at 20:14
  • @Student If you chase the links I gave, you will find an answer where I show how to simply and mechanically derive the form of the proof given in the book using the viewpoint I described above. And, to boot, I show how to do it in divisiblity form too (for readers who don't yet know congruences). But one should not aim to derive proofs that obfuscate the essence of the matter. Rather one should aim to derive proofs that highlight the essence. – Bill Dubuque Jul 14 '17 at 20:19
  • @AntonRasmussen: You are free to change your mind and accept a different answer, if you wish. That would not indicate that the original accepted answer is bad, but just indicates which one you felt helps you the most. You can upvote every answer that helped you. – user21820 Jul 18 '17 at 10:21
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I don't think there is an algorithm for that, but I'd start by simplifying $$ 7^{n+2} + 8^{2n+1} = 49\cdot 7^n + 8 \cdot 64^n $$ Now note that $49+8=57$ and $64-7=57$, which at least suggests where $57$ comes from.

This observation is probably useful if you write $64^n=(57+7)^n$ and use the binomial theorem.

lhf
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\begin{align} 7^{n+2}+8^{2n+1}&=49\cdot 7^n+8\cdot 64^n \\ &\equiv -8\cdot 7^n+8\cdot 7^n\equiv 0\bmod{57}. \end{align}

Paolo Leonetti
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Hint $ $ To induct simply multiply the first two congruences below (by the Congruence Product Rule)

$$\begin{align} \bmod 57\!:\qquad\ \ 7&\equiv 8^{\large 2}\\[.3em] {-}7^{\large n+2}&\equiv 8^{\large 2n+1}\ \ \ {\rm i.e.}\ \ P(n)\\ \overset{\rm multiply}\Longrightarrow\ {-}7^{\large n+3} &\equiv 8^{\large 2n+3}\ \ \ {\rm i.e.}\ \ P(n\!+\!1) \end{align}$$

Remark $ $ Just as above, the arithmetical essence of many inductive proofs is greatly clarified using congruence language. In fact we can make it clearer by showing that the induction amounts to raising $\,7\equiv 8^2$ to power $n,\,$ using the Congruence Power Rule, then multiplying the result by the base-case congruence $\,-7^2\equiv 8,\,$ using the Congruence Product Rule, $ $ i.e.

$$\qquad \qquad\qquad \begin{align} \bmod 57\!:\quad {-}7^2&\equiv 8\qquad\quad\ \rm [Base\ Case]\\[.3em] 7^{\large n}&\equiv 8^{\large 2n}\ \ \ {\rm by}\ \ [7\equiv 8^{\large 2}]^{\large n}\ \ {\rm by\ \ Power\ Rule}\\ \overset{\rm multiply}\Longrightarrow\ {-}7^{\large n+2} &\equiv 8^{\large 2n+1} \end{align}$$

Thus we see that the arithmetical essence of the induction on $\,n\,$ amounts to the fact that congruences are preserved under raising to power $\,n,\,$ whose inductive proof has been abstractly encapsulated into a conveniently reusable lemma, the Congruence Power Rule. Once we are familiar with this inductive pattern, we can easily recognize when it occurs in future inductions, eliminating the often difficult meandering search for the key idea for the inductive step.

Furthermore, using the above viewpoint we can mechanically generate the proof in your book, e.g. see this answer where I do that explicitly for two proofs. However, we should not aim to derive proofs that obfuscate the (arithmetical) essence of the matter. Rather, we should aim to derive proofs that highlight the essence - as above.

Bill Dubuque
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By the Chinese remainder theorem something is $\equiv 0\pmod{57}$ iff it is $\equiv 0\pmod{3}$ and $\equiv 0\pmod{19}$. On the other hand $$ a_n = 7^{n+2}+8^{2n+1} = 49\cdot 7^n + 8\cdot 64^n $$ fulfills $$ a_n \equiv 49\cdot 1^n+8\cdot 1^n \equiv 57 \equiv 0\pmod{3} $$ $$ a_n \equiv 49\cdot 7^n+8\cdot 7^n \equiv (-8+8)7^n \equiv 0\pmod{19} $$ hence you do not need induction.

Jack D'Aurizio
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The steps to the third line seem routine, trying to find the terms of the inductive hypothesis. Once you are at the third line you have to decide to split the $64$ into $7+57$. You might just notice that both numbers are important in the problem and try it. You might notice that splitting out $7$ of the second term is what is needed to complete the inductive hypothesis in the first term, so try it. When you see $57$ is left you should be convinced that is the right track.

Ross Millikan
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"..for every non-negative integer $n$" means that it shall be valid starting from $n=0$.
In fact $F(0)= 7^2+8=57$.

You already found that $F(n+1)=7 \,F(n)+57\,8^{2n+1}$ which is clearly divisible by 57.
So $$57\backslash F(n)\quad \left| {\;0 \le n} \right.$$

G Cab
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you would have to think about how to get the form desired. in this case the first part of:

$7(7^{k+2}+8^{2k+1})+57⋅8^{2k+1}$

We have made a factor, of the form desired. Assuming it's divisible by 57, that part of the sum is, and the other part shows it is already. So, the sum of both parts, must divide by 57. Hence, the original number must divide by 57. The original number is of the form suggested but with n replaced by k+1. so it must be true for n=k+1, assuming it works for n=k. then what's left this to show it works for some base value of k. As $7^3+8^3=343+512=855=57\cdot {15}$ it works for n=k=1, and it follows by what we've shown that it then works for all k>1 . edit: okay I forgot the 0 case ( see G cab's answer): as 49+8=57, we also have it for the k=0 case, and again all other cases are shown by induction.