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I have the concepts of induction well-understood. But even so, solving these kind of problems isn't that easy.

Please, Someone show me how to do the first one, and explain a bit of how I can have vision to solve problems like this. The second one I'll try by myself based on the first.

I tried to do the first example from what I already knew, but I failed.

a) $$3 | n(7+6n+2n^2)\quad \forall n \in \mathbb{N} $$

Ok, first step is to plug $1$ to check if it works for the first $n \in \mathbb{N}$. It's ok.

Now I need to plug $k$ and $k+1$, to make sure any value $k$ and its consecutive one will do for this statment.

$$k(7+6k+2k^2)$$

$$(k+1)(7+6(k+1)+2(k+1)^2)$$

Now I need to have a good vision to be able to find $k(7+6k+2k^2)$.

b) $$6|(n^3 + 5n) \quad \forall n \in \mathbb{N} $$

Edit: Bill Dubuque's suggetion

$$ 6|6d-k^3-5k = f(k)$$ $$Then$$ $$ 6|6j -(k+1)^3 -5(k+1)=f(k+1)$$

$$<==>$$ $$6|6j -(k+1)^3 -5(k+1) - (6d-k^3-5k)$$

$$\quad \forall d \in \mathbb{N} $$ $$\quad \forall j \in \mathbb{N} $$

Goun2
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3 Answers3

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It is simply a matter of noticing that\begin{align*}(k+1)\bigl(7+6(k+1)+2(k+1)^2\bigr)-k(7+6k+2k^2)&=15+18k+6k^2\\&=3\times(5+6k+2k^2).\end{align*}

José Carlos Santos
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  • i.e. $,f(k+1)-f(k),$ is divisible by $3,,$ which amounts to using telescopic induction - see my answer. – Bill Dubuque Jul 08 '17 at 21:47
  • I did the second one as you did in the first one, and it worked, I didn't even need to pay attention to details in order to get the result, by far the the less complicated way. – Goun2 Jul 09 '17 at 01:47
  • Does that way work for every problem ?, I tried to do the same with another example and it didn't work. 6|5^k + 2*11^k. – Goun2 Jul 09 '17 at 03:26
  • @Goun2 $6$ doesn't divide $5^k + 2 \times 11^k$, since odd + even = odd but $6$ is even. Maybe you meant $3 | 5^k + 2\times 11^k$, in which case normal induction will do, i.e. if $5^k + 2\times 11^k = 3m$ for some $m$, then $5^{k+1} + 2\times 11^{k+1} = 5 (5^k + 2\times 11^k) + 2 \times 6 \times 11^k = 3(5 m + 4 \times 11^k)$. – Weaam Jul 09 '17 at 04:34
  • @Weaam I had written wrong, It is 3 in fact. But I wanted to check if that problem could be solved as the accepted answer. (5^(k+1)+211^(k+1)) - (5^k + 211^k). – Goun2 Jul 09 '17 at 05:13
  • @Goun2 It can be, but using a modular arithmetic argument, i.e. since $5, 11 \equiv 2 \mod 3$ then $2^k (2-1) + 2\times 2^k (2-1) = 2^k - 2^k \equiv 0 \mod 3$ since $2 \equiv -1 \mod 3$. – Weaam Jul 09 '17 at 05:39
  • @Goun2 Try not to rely on a single method. Usually a problem will be easier to solve with one method over the other. So it's important to have a more diverse set of tools. – Weaam Jul 09 '17 at 05:44
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1) for $n=1$; (7 + 6 + 2) = 15 = 3$\times$5; ok!

Assuming it is valid for $n$, then $n(7+6n+2n^2) = 3k$, for some $k$ $\in$ $\mathbb{N}$, and \begin{align*}(n+1)(7 + 6(n+1) + 2(n+1)^2) &= n(7 + 6(n+1) + 2(n+1)^2) + (7+ 6(n+1) + 2(n+1)^2 )\\ &= n(7 + 6n + 2n^2) + (6n + 4n^2 + 2n) + (7+ 6(n+1) + (2(n+1)^2)\\ &= 3k + 8n + 4n^2 + 13 + 6n + 2n^2 + 4n + 2\\ &= 3k + 6n^2 + 18n + 15 = 3(k + n^2 + 6n + 5) \end{align*}

$$\Rightarrow 3| (n+1)(7 + 6(n+1) + 2(n+1)^2)$$

2) for $n=1; 1 + 5 = 6$; ok!

Assuming it is valid for $n$, then $n^3 + 5n = 6k$, for some $k$ $\in$ $\mathbb{N}$, and

\begin{align*} (n+1)^3 + 5(n+1) &= (n^3 + 5n) + (3n^2 + 3n + 1 + 5)\\ &= 6k + (3n^2 + 3n + 6) \\ &= 6k + 3(n^2 + n + 2) \\ &= 6k + 3(n(n+1) + 2 )\\ \end{align*} as $n(n+1)$ and $2$ are even, then $n(n+1) + 2 = 2w$, for some $w$ $\in$ $\mathbb{N}$ $$\Rightarrow (n+1)^3 + 5(n+1) = 6k + 3(2w) = 6(k+w)$$.

$$\Rightarrow 6| (n+1)^3 + 5(n+1)$$

Matheus Manzatto
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  • Worth emphasis: you essentially proved that $,f(n+1)-f(n),$ is divisible by $3,,$ which amounts to using telescopic induction - see my answer. – Bill Dubuque Jul 08 '17 at 21:44
  • You shortened your answer a bit and I couldn't follow, could you give more details about the first one ? I only understand the first line, where you multiply by 1 and add and leave only the n multiplying. – Goun2 Jul 13 '17 at 05:31
  • On the second line, I used that $n(7+6(n+1)+2(n+1)^2) = n (7 + 6n + 6 + 2n + 2 + 2n^2 + 4n + 2) $ $=$ $ n ( (7 + 6n + 2n^2) + (6 + 4n +2) ) = n(7 + 6n + 2n^2) + (6n +4n^2 + 2n) $. On third line I used the induction hypotesys on $n(7 + 6n + 2n^2)$, what give to us that $n(7 + 6n + 2n^2)$ is a multiple of 3, so .$n(7 + 6n + 2n^2)= 3k$, for some $k$ – Matheus Manzatto Jul 13 '17 at 05:57
  • To reach until the fourth line the only thing I did was to expand the equation, I just did the calculations and you can verify this result doing the calculations as you want, after this come the important part, I put the number 3 in evidence, what show that (n+1)(7+6(n+1)+2(n+1)2)(n+1)(7+6(n+1)+2(n+1)2) is a multiple of 3, what verificate the induction hypothesis. Hence 1) e 2) we conclude that all natural number with the form n(7+6n+2n^2) are divisible by 3 – Matheus Manzatto Jul 13 '17 at 05:58
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    @MatheusManzatto Thank you bro. – Goun2 Jul 13 '17 at 06:01
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If $\,d\mid f(n)$ then $\,d\mid f(n+1)\iff d\mid f(n+1)-f(n),\,$ so it suffices to prove the latter to prove the induction step. Doing so is mechanical polynomial arithmetic which proves it true (this is essentially what is done in the other two answers). This is a special case of telescopic induction. You can find much further discussion of this powerful method in prior posts on telescopy.

Or, exploiting modular arithmetic reveals innate arithmetical structure which simplifies things.

$\qquad\bmod 3\!:\,\ n(7+6n+2n^2)\equiv n(1-n^2)\equiv -(n-1)n(n+1)$

and it is easy to prove that a sequence of $k$ consecutive integers contains a multiple of $k$ (either by explicit induction, or otherwise, e.g. by the division algorithm, etc).

Same for the 2nd problem since $\,n^3+5n\equiv n^3-n\pmod{6},\,$ same as above, so being a product of a sequence of three consecutive integers it contains a multiple of $2$ and $3,\,$ so their product is a multiple of $\,{\rm lcm}(2,3) = 6$.

Bill Dubuque
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  • Wish there's a double up-vote option for such answers. Succinct and quite informative. Thank you! – Weaam Jul 09 '17 at 00:02
  • I started learning modular arithmetic yesterday, I only know the initial concepts of M.A, What is the name of that propriety you showed ?, your answer wouldn't be accpeted in my exame because I really need to prove as the others answers, But I really aprecciate to know how things work in detail so it becomes easier to understand. – Goun2 Jul 09 '17 at 00:36
  • @Goun2 The property in the proof hinted in the first paragraph (that is implicitly used in the other answers) does not have a standard name. I refer to it as telescopic induction (see answers in the final link). – Bill Dubuque Jul 09 '17 at 15:33
  • @BillDubuque There aren't many resources to learn this subject on the internet, could you recommend some book ? – Goun2 Jul 09 '17 at 22:12
  • @BillDubuque I tried to do the second one by the telescopic induction, could you please check if it is right ? – Goun2 Jul 10 '17 at 14:17
  • @Goun2 For that we get $,f(n+1)-f(n) = 3(n(n+1)+2)$ which is divisible by $6$ since $g(n) = n(n+2)+2$ is even (since $n$ or $n+1$ is even). It amounts to the same proof in Matheus' answer. Or you could do the same for $g$, i.e. $g(0)=2$ is even and $g(n+1)-g(n) = 2(n+1)$ is even, so $g(n)$ is even for all $n$ by telescopic induction. – Bill Dubuque Jul 10 '17 at 15:36
  • @BillDubuque I think i'm lacking some prerequisite knowledge to understand you, As I got interested in learning this telescopy induction I'll look for it in some book, anyways thank you for your attention. – Goun2 Jul 10 '17 at 22:59
  • @Goun2 If you let me know what is proving difficult I will be happy to elaborate. Unfortunately it is not discussed in any books that I am aware of. There are many other examples in the posts I linked above. – Bill Dubuque Jul 11 '17 at 00:50