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I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?

$$\begin{align} 16^m - 1=&\frac{16^{m+1}}{16}-1\\ =&\frac{16^{m+1}-16}{16} \\ =&(16^{m+1}-16)\cdot\frac{1}{16} \\ =&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\ =&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\ =&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\ =&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\ =&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\ =&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\ \end{align}$$ $$\therefore \boxed{16^m - 1=\frac{b}{15}}$$

Or is this the wrong way and I have to do it with mathematical induction?

gosua
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  • You have to use induction since, apparently you can't or still haven't studied modular arithmetic (way easier). – DonAntonio Nov 16 '13 at 20:15
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    Your approach is flawed, you've essentially shown that $b=15(16^n-1)$ exists. –  Nov 16 '13 at 20:19
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    This proof is not right and I explained it in my answer but, just curious, what caused you to check your proof? Why did you doubt the proof? What made you suspicious? – hhsaffar Nov 16 '13 at 20:31
  • Related : http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b and http://math.stackexchange.com/questions/497235/how-to-show-that-10n-1-is-divisible-by-9 – lab bhattacharjee Nov 17 '13 at 05:17
  • @hhsaffar : because we regularly proofed such kind of tasks in the way that we formed the expression to a multiplication e.g. a*15 and this lead us to the fact, that a is divisable by 15. my "flawed" approach above didn't had that form and so this made me suspicious that it could be wrong. – gosua Nov 17 '13 at 07:31

5 Answers5

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Hint:

$$16^{n+1}-1=16\cdot 16^n-1=15\cdot 16^n+(16^n-1)\;\ldots$$

DonAntonio
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HINT:Use induction and $$16^{n+1}-1=16(16^n-1)+15$$

Adi Dani
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$$ 16^{n} - 1 = 15\,{16^{n} - 1 \over 16 - 1} = 15\sum_{k = 0}^{n - 1}16^{k} $$ $$ \color{#0000ff}{\large{16^{n} - 1 \over 15} = \sum_{k = 0}^{n - 1}16^{k}} $$

Felix Marin
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Just observe that $x^n-1$ factors as $(x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)$, and make the appropriate substitution.

Doc
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Actually you have shown that there is a number $b$ that if you divide it by $15$ you will get $16^n-1$. But this does not mean that $16^n-1$ is divisible by $15$. For example $23=\frac{345}{15}$ but it does not mean that $23$ is divisible by $15$ . To show that $a$ is divisible by $b$ you have to show that $a=kb$ ($k$ is an integer, just like $a,b$)not $a=\frac kb$. Adi Dani Hinted for a solution by induction.

hhsaffar
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