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I have a question. I must prove with induction that $$\displaystyle 2^{2n+1} - 9n^{2} + 3n - 2$$ is divisible by $54$.

My attempt was that if I plug in $n = 0$, I get result of $0$.

Then I want to show this sentence for $n + 1$:

$2^{2(n+1) + 1} - 9(n + 1)^2 + 3(n + 1) - 2 = 2^{2n + 3} - 9n^2 - 18n - 9 + 3n + 3 - 2 = (2^{2n + 1} \cdot 4 - 9n^2 + 3n - 2 ) - 18n - 6 = (3 \cdot 2^{2n + 1} - 18n - 6) + \left(2^{2n + 1} - 9n^{2} + 3n - 2\right)$.

Here I stopped, because I don't know how to manipulate the prove. Can anyone help me?

Mike
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Andrej Benedičič
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  • I think induction is the right idea, but you may want to use the first few terms to make the manipulation easier. You could also split up the 54 and show the expression is divisible by 2 and then 27. At least until you've proved it to yourself. – Leonhard Euler Jun 05 '21 at 23:29
  • Your first term is obviously divisible by 6. So you can take that out as a factor. It remains to show the rest is divisible by 9. – Leonhard Euler Jun 05 '21 at 23:32
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    I added another answer to the linked dupe to show how it follows easily from the Bionomial Theorem (first three terms). This is a common way to simplify such inductions, i.e. they are essentially special cases of the inductive proof of the Binomial theorem, modulo minor transformations (here doubling to get twice the modulus). – Bill Dubuque Jun 06 '21 at 01:30

1 Answers1

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Based on your work so far, it suffices to show $3 \cdot 2^{2n+1} - 18 n - 6$ is divisible by $54$.

By applying the inductive hypothesis again, we know $2^{2n+1} \equiv 9n^2 - 3n + 2 \mod 54$ so $$3 \cdot 2^{2n+1} - 18 n - 6 \equiv 3 \cdot (9n^2-3n+2) - 18n - 6 \mod 54$$

The right-hand side simplifies to $27n(n-1)$ which is divisible by $54$.

angryavian
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