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Currently I am on mathematical induction and I've faced problem that I simply don't know where to even start and I can't find any examples that I can go on with, so just to clarify I am asking you for example with solution of similar task.

The task that I should solve is: $$1053|3^{2n+2}*5^{2n}-3^{3n+2}*2^{2n}$$

If you possibly give me solution of this: $$3|(n^{3}-n)$$ since I believe this one is easier to solve

P.S: I don't want you to solve my problem, because I want to try to solve it by myself, I just want an example to see how you can solve these type of tasks. Thank you!

str1ng
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    Suggested reading: How to write a clear induction proof? – JMoravitz Sep 24 '20 at 18:05
  • @JMoravitz Yeah but this type of writing induction proof doesn't have to do anything with mine task. Even in the workbook it's given like this – str1ng Sep 24 '20 at 18:07
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    For your smaller example, the induction step will look something like $(n+1)^3 - (n+1) = \cdots = (n^3-n)\cdot a + 3\cdot b$ with $a$ and $b$ some integers (possibly depending on $n$) that you arrive at after algebraic manipulation, thus $(n+1)^3-(n+1)$ is equal to something times $(n^3-n)$ which we know is a multiple of three (per our induction hypothesis) plus another multiple of three and is thus also a multiple of three. – JMoravitz Sep 24 '20 at 18:08

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Use this: $$n^3-n=(n-1)n(n+1)$$ or (if you want a proof by induction): $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=n^3-n+(3n^2+3n).$$

Michael Rozenberg
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  • So basically as if I am proving induction with " = " ; I take variables and go through all 3 steps ?

    And I am assuming that I have to prove that right side is same like the number on the left side?

     – str1ng Sep 24 '20 at 18:09
  • @Pshyotic You need to check that $n^3-n$ is divisible by $3$ for $n=1$. Also, you need to prove the following statement: $n^3-n$ is divisible by $3$ $\Rightarrow$ $(n+1)^3-(n+1)$ is divisible by $3$. The last statement follows from the equality: $(n+1)^3-(n+1)=n^3-n+3n(n+1).$ – Michael Rozenberg Sep 24 '20 at 18:13
  • Well then it's basically proved by that, isn't it? – str1ng Sep 24 '20 at 18:16
  • @Pshyotic Yes, by the axioma of an induction. – Michael Rozenberg Sep 24 '20 at 18:17
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    Thank you a lot :) I could understand pretty much most of it. The other guy helped me clarify it – str1ng Sep 24 '20 at 18:25
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This following problem is more similar to your $1053$ problem: show that $5 \mid (3^{2n}-2^{2n})$.

The base case is trivial.

For the inductive step, we assume that $5 \mid (3^{2k}-2^{2k})$ and try to prove that $5 \mid (3^{2(k+1)}-2^{2(k+1)})$.

We write $3^{2k}-2^{2k}=5K$, so we have $3^{2k} = 5K+2^{2k}$. Now:

$$\begin{align}3^{2(k+1)}-2^{2(k+1)}&=9(5K+2^{2k})- 2^{2(k+1)}\\&=45K+9\cdot2^{2k}-4\cdot2^{2k}\\&=5(9K+2^{2k})\end{align}$$

which shows our desired result.

This is a standard method for "divisibility of powers type" mathematical induction proofs.

player3236
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Hint: for the inductive step, use $(n+1)^3 - (n+1) = n^3 + 3 n ^2 + 3 n + 1 - n - 1 = (n^3 - n) + 3 (n^2 + n) $ . What can you say for each of these two terms?

Javi
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  • As far as I can see we have three consecutive numbers, which means that one of them must be even and all are divisible by 3 – str1ng Sep 24 '20 at 18:14
  • im not following you. What three numbers do you mean? And how can three consecutive numbers be all divisible by 3? – Javi Sep 24 '20 at 18:18
  • Well $$n^{3} - n = n(n^{2}-1^{2}$$ you can clearly see n+1 ; n ; n-1 Those are 3 consecutive numbers and if I am not wrong when you have 3 consecutive numbers they must be even and divisible by 3 – str1ng Sep 24 '20 at 18:21
  • at least one of them is even and one is odd... and one and only one of them is divisible by 3. But we are going away from what you want to do... can you apply the steps for induction? – Javi Sep 24 '20 at 18:26
  • Sorry, because I wasn't active previous 2 days, I succesfully solved my task and now I can understand how you can solve tasks like this. Thank you very much on your help – str1ng Sep 27 '20 at 11:00