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This question is about an inductive proof which was posted yesterday on this web site here: https://math.stackexchange.com/questions/1371540/proving-frac5-cdot34n-1-22n7-is-an-integer. This topic was put on hold as off topic. I'm pretty rusty on algebra and just getting back in the game. Can someone please explain how the algebra works in the last part of the step in red. I see that one $9^{2n}$ was taken from $15\cdot9^{2n}$ so that this term is divisible by $7$ ($14 \cdot 9^{2n}$ is divisible by $7$). I also see that the terms which was added after $14 \cdot 9^{2n} ((9 - 2) \cdots$) are there to "fix things up" and keep the equation the same. However, I don't see how the fix works. Could someone please explain how this works.

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Willard
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    The practice of using the phrase "off topic" to refer to things like that posting is rude, but standard on this site. ${}\qquad{}$ – Michael Hardy Jul 24 '15 at 14:34
  • I'm certainly not trying to be rude! I'm just a guest here and was only pointing out the actual state of the question I linked to. Thank you for answering my question. – Willard Jul 24 '15 at 14:39
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    It wasn't you I was calling rude; it was those who use that phrase in this context. You were only quoting them. ${}\qquad{}$ – Michael Hardy Jul 24 '15 at 14:46

2 Answers2

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$15\cdot 9^{2n} - 2^{2n} $

$(14 + 1) \cdot 9^{2n} - 2^{2n} $ Since 15 = 14 +1

$14 \cdot 9^{2n} + 9^{2n} - 2^{2n} $ By distributive law

$14 \cdot 9^{2n} + ( 9 - 2) \sum_{i=1}^{2n} 9^{i} 2^{2n - i}$ It is factoring $9^{2n} - 2^{2n}$

$14 \cdot 9^{2n} + ( 7) \sum_{i=1}^{2n} 9^{i} 2^{2n - i}$

To see how to factor the last thing see this http://www.mathwords.com/f/factoring_rules.htm

IrbidMath
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$$ 5 \cdot 3^{4n+1}-2^{2n} = 15\cdot 9^{2n}-2^{2n} = \color{red}{14}\cdot 9^{2n}+\color{red}{(9-2)}\cdot\sum_{k=1}^{2n}2^{k-1} 9^{2n-k}. $$

$$ 15\cdot 9^{2n}-2^{2n} = (14+1)9^{2n} - 2^{2n} = 14\cdot 9^{2n} +\Big( 9^{2n} - 2^{2n} \Big) $$ The problem now is to show that $9^{2n} - 2^{2n}$ is divisible by $7$, i.e. by $9-2$. Here's a concrete instance: $$ 9^8 - 2^8 = (9-2) \Big( 9^7 + (9^6\cdot2) + (9^5\cdot2^2) + (9^4\cdot 2^3) + (9^3\cdot2^4) + (9^2\cdot 2^5) + (9\cdot 2^6) + 2^7 \Big). $$ The last factor can be written as $\displaystyle\sum_{k=1}^8 9^{8-k} \cdot 2^{k-1}$.

If you multiply out the concrete instance, just first multiply the whole expression in the $\Big(\text{big parentheses}\Big)$ by $9$, then by $-2$, and observe that every postive term except the first one cancels out one of the negative terms, and every negative term except the last one cancels out one of the positive terms. You're left with the first postive term and the last negative term, i.e. with $9^8 - 2^8$.

Michael Hardy
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