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I know that this question was already answered, but I would like to know if the second step of induction its okay the way I did it.

This question is different from Induction proof for $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$ and Let $n ∈ N, n ≥ 1$. Prove that $4^n + 6n - 1$ is divisible by $9$. since it is just about to check a part from a proof that I did (if the way I thought is correct or not).

Suppose that $9 \mathrel| (4^k+6k-1)$ is true for $k\in \mathbb{N}, k\geq1$. I want to prove $P(k+1)$: $ 9 \mathrel| 4^{(k+1)} +6(k+1) - 1$

Since $9 \mathrel| (4^k+6k-1)$, I can rewrite as:

$4^k+6k-1 = 9q$, where $q \in \mathbb{Z} $. By the definion of Divisibility.

  • Multiplying the equation on both sides by 4, I'll get:

$4^{(k+1)} +24k-4 = 36q$

  • Adding 3 on both sides

$4^{(k+1)} +24k - 1 = 36q + 3$

  • Subtracting $18k$ on both sides

$4^{(k+1)} +6k - 1 = 36q + 3 -18k$

  • Adding 6 in order to have $6(k+1)$ as a factor

$4^{(k+1)} +6k + 6 - 1 = 36q + 3 -18k + 6$

$4^{(k+1)} +6(k+1) - 1 = 36q + 9 -18k $

Note that $36q + 9 -18k$ can be written like this: $9(4q + 1 - 2k)$. Thus $9Q$, $Q=(4q + 1 - 2k), Q \in \mathbb{Z}$.

Therefore:

  • $4^{(k+1)} +6(k+1) - 1 = 9Q$. Which means that $4^{(k+1)} +6(k+1) - 1$ is also divided by 9.

Does my proof look fine? If there is a problem, please tell me. All help is appreciated.

Daniel Sehn Colao
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    you haven't started the induction step with $P(k+1)$, so no prize for you – AgentS Aug 17 '19 at 23:11
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    It's much quicker with congruences. – Bernard Aug 17 '19 at 23:14
  • I missed it @ganeshie8. Thanks. – Daniel Sehn Colao Aug 17 '19 at 23:15
  • It works. You got were you were driving through. Adding $3$ to both sides set my panic button off because we could not longer claim both sides were divisible by $9$ but that wasn't where you aiming the steering wheel after all so it was all good in the end... people point out you can make it shorter and neater but I like that I can see you worked this through on your own. – fleablood Aug 18 '19 at 00:12
  • Yeah @fleablood. Thanks. I made a lot of comments during the proof even though some were unnecessary, but I did just in order to make it clearly and understandable. – Daniel Sehn Colao Aug 18 '19 at 00:28

5 Answers5

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There is an easier way: $4^n+6n-1\iff (3+1)^n+6n-1=(9M+3n+1)+6n-1=9(M+n)$

Piquito
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It's fine. Cut to the essentials: just say if $P(k)=9q$ for some $q$ then $P(k+1)=9(4q-2k+1)$, with some minimal computations to justify it:

$$P(k+1)=4^{k+1} + 6(k+1)-1 = 4(4^k) + 6k +5 =\\ 4(9q-6k+1)+ 6k+5 = 36q - 18k + 9 = 9(4q-2k+1)$$

No "cleverness" needed, it just falls out automatically.

Henno Brandsma
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  • Or, that the difference $3\cdot 4^k+6$ must divide by 9 leading to $4^k$ always having remainder 1 on division by 3. which is true, proving that the differences are, and so if one element of the original divides by 9, they all do. –  Aug 17 '19 at 23:26
  • @RoddyMacPhee I think the OP wants a simple computation style answer as I gave it, no congruences etc. – Henno Brandsma Aug 17 '19 at 23:29
  • you can turn mod into linear polynomials. –  Aug 17 '19 at 23:39
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It's good. I like it a lot.

But in induction proofs it's often easier to start at the destination and drive backwards through one way streets to the start, rather than starting from the start and driving through a bunch of intersections without a map in order to reach your destination.

$4^{k+1} + 6(k+1) - 1=$

$4*4^{k} + 6k + 5 = $

$4(4^k + 6k -1) - 24k +4 +6k + 5=$

$4(4^k + 6k -1) - 18k +9 = 4(4^k+6k-1) -9(2k-1)$.

And $9|4^k + 6k -1\implies 9|4(4^k+ 6-1) - 9(2k-1)$$

fleablood
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If $n \equiv 0 \pmod 3$, then for an integer $k$:

$$4^n + 6n-1 \equiv 4^{3k} + 18k - 1 \equiv 1^k + 0 - 1 \equiv 0 \pmod 9.$$

If $n \equiv 1 \pmod 3$, then:

$$4^n + 6n-1 \equiv 4 \cdot4^{3k} + 18k + 6- 1 \equiv 4+0+6-1 \equiv 0 \pmod 9.$$

Finally, if $n \equiv 2 \pmod 3$, we have:

$$4^n + 6n-1 \equiv 16 \cdot4^{3k} + 18k + 12- 1 \equiv 16+0+12-1 \equiv 0 \pmod 9.$$

Since an integer can only congruent to either $0 \pmod 3, 1 \pmod 3$ or $2 \pmod 3$, this completes the proof for all integer $n$.

Toby Mak
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It is fine. The same work with congruence: $$4^k+6k-1\equiv 0 \pmod{9}$$ Since $gcd(4,9)=1$, multiply it by $4$: $$4^{k+1}+24k-4\equiv 4^{k+1}+\require{cancel}\cancel{18k}+6k-4+\color{red}9\equiv 4^{k+1}+6(k+1)-1\equiv 0\pmod{9}.$$

farruhota
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