Prove without using induction that all numbers of the form $6|8^n - 2^n$. I need a brush up on subtracting numbers with the same base but different exponent. So far I have $8^n - 2^n = 2^{3n} - 2^{n}$. Am I headed in the right direction?
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Hint: $$8^n - 2^n = (8 - 2)(8^{n-1} + 8^{n-2}2 + \cdots + 8 \ 2^{n-2} + 2^{n-1})$$
Ivo Terek
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I think this is the way to think about this problem, reverting to the special fact that $2^3=8$ only if required. Apart from anything else $y-x$ is a factor of $y^n-x^n$ makes this easy to see with no computation. – Mark Bennet Oct 28 '14 at 22:12
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HINT
$8^n - 2^n=(2^n-1)\cdot 2^n\cdot (2^n+1)$
georg
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This is exactly what I was about to post. This is surely the most obvious reason. +1 for you -- great minds think alike ;) – MPW Oct 28 '14 at 21:25
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Thank you. Please correct me if I'm wrong, but $8^n - 2^n = 2^{3n} - 2^{n} = 2^n(2^{2n} - 1) = 2^n(2^n - 1)(2^n + 1)$ – mylasthope Oct 28 '14 at 21:47
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@MPW :-) mylasthope: I would say $8^n - 2^n = 2^{3n} - 2^{n} = 2^n(2^{2n} - 1) = 2^n(2^{n} - 1)(2^n + 1)$ – georg Oct 28 '14 at 21:56
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$8^n-2^n = 2^n(4^n-1) = 2 \cdot 2^{n-1} \cdot 3 \cdot (4^{n-1} + \cdots + 4+1)$
Crostul
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