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So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.

Basic Step: Let $n = 1$. Therefore,

$$ 9^{1+1} + 7^{2 \cdot 1} = 9^2 + 7^2 = 130 $$

Therefore, $10|(9^{n+1}+7^{2n})$, is true.

Inductive Step: There exists an integer $k$ such that $(9^{n+1}+7^{2n}) = 10k$. Let $P(k) = 9^{k+1} + 7^{2k}$. Therefore,

$$ P(k+1) = 9^{(k+1)+1} + 7^{2(k+1)} = 9^{k+2} + 7^{2k+2} $$

$$ P(k+1) = (9^1 \cdot 9^{k + 1}) + (7^2 \cdot 7^{2k}) $$

$$P(k+1) = (2+7)(9^{k+1}) + (7^2 \cdot 7^{2k}) = (2 \cdot 9^{k+1}) + (7 \cdot 9^{k+1}) + (7 \cdot 7 \cdot 7^{2k}) $$

$$ P(k+1) = (2 \cdot 9^{k+1}) + 7(9^{k+1} + 7 \cdot 7^{2k}) $$

J. W. Tanner
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DH00325
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  • $9^{n+1}+7^{2n}\equiv(-1)^{n+1}+(-1)^n\equiv0\bmod10$ – J. W. Tanner Oct 25 '20 at 01:26
  • Use https://math.stackexchange.com/questions/3860386/i-am-stuck-in-this-question-proof-by-induction-divisibility/3860401?r=SearchResults#3860401 or https://math.stackexchange.com/questions/3486390/is-it-valid-to-subtract-in-the-middle-of-the-proof-here/3486393?r=SearchResults#3486393 – lab bhattacharjee Oct 25 '20 at 02:36

3 Answers3

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Can you show that $$9^{k+2}+7^{2k+2}=9\cdot9^{k+1}+49\cdot7^{2k}=9(9^{k+1}+7^{2k})+40\cdot7^{2k}$$ is divisible by $10$?

J. W. Tanner
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Instead of trying to factor out numbers to make it divisible by 10, I was able to solve the equation by rearranging the formula that equals 10k to

$$ 9^{+1} = 10 - 7^{2} $$

If you substitue it into P(k + 1) you get:

$$ P(k+1) = 9^1 \cdot (10m - 7^{2k}) + (49 \cdot 7^{2k}) $$ Which reduces to $$ P(k+1) = 90m + (40 \cdot 7^{2k}) = 10(9m + (4 \cdot 7^{2k}) $$ which makes it divisible by 10

DH00325
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By the binomial theorem, $$9^{n+1}+7^{2n}=(10-1)^{n+1}-(50-1)^n=(10a+(-1)^{n+1})-(50b+(-1)^n)=10a-50b$$ since $n$ and $n+1$ have different parities.

lhf
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