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Without using induction, how can it be proved that 7 divides $3^{2n+1}+2^{n+2}$ for each $n\in\mathbb{N}$? I tried to expand it using $\frac{x^{n+1}-1}{x-1}=1+x+..+x^n$ but I had no success. It would be great if more than one proof is provided.

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  • $\begingroup$ What have you tried? $\endgroup$
    – pancini
    Aug 18, 2020 at 22:50
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    $\begingroup$ It would be great if even one attempt were provided! What have you tried? $\endgroup$
    – user208649
    Aug 18, 2020 at 22:50
  • $\begingroup$ I have edited . Great sense humor ahaha @TokenToucan. $\endgroup$
    – user723846
    Aug 18, 2020 at 22:53
  • $\begingroup$ Any number other than zero divides zero! $\endgroup$ Aug 18, 2020 at 22:54
  • $\begingroup$ In plain language, the quantity under consideration is zero modulo 7! $\endgroup$ Aug 18, 2020 at 23:13

4 Answers 4

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\begin{eqnarray*} \sum_{n=0}^{\infty} (3^{2n+1}+2^{n+2})x^n = \frac{3}{1-9x}+\frac{4}{1-2x} = \frac{ \color{red}{7} (1-6x)}{(1-9x)(1-2x)}. \end{eqnarray*} This function clearly has integer coefficients \begin{eqnarray*} \frac{ (1-6x)}{(1-9x)(1-2x)}=(1-6x) \left( 1 +9x+81x^2+ \cdots \right) \left( 1 +2x+4x^2+ \cdots \right). \end{eqnarray*}

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  • $\begingroup$ Do not you mean $\frac{2}{1-2x}$ instead of $\frac{4}{1-2x}$? $\endgroup$
    – user723846
    Aug 18, 2020 at 23:08
  • $\begingroup$ It is perfect... but how can I deduce that 7 divides each term of the series? $\endgroup$
    – user723846
    Aug 18, 2020 at 23:20
  • $\begingroup$ Great solution! $\endgroup$
    – user723846
    Aug 18, 2020 at 23:26
  • $\begingroup$ Thanks @Gio $ \ddot \smile$ $\endgroup$ Aug 18, 2020 at 23:27
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HINT: Simplify $3^{2n+1}+2^{n+2}$ modulo $7$, using the fact that $3^{2n+1}=3\cdot 3^{2n}$ and $3^2\equiv2\pmod7$.

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    $\begingroup$ Was just about to write this too $\endgroup$
    – CryoDrakon
    Aug 18, 2020 at 22:55
  • $\begingroup$ for this method, would you have to do case work? $\endgroup$
    – C Squared
    Aug 19, 2020 at 0:36
  • $\begingroup$ @CSquared: No; it’s a one-line proof. $\endgroup$ Aug 19, 2020 at 0:46
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$3^{2n + 1} + 2^{n+2} = 3\cdot 3^{2n} + 2^2\cdot 2^n = 3\cdot(9)^n + 4\cdot s2^n\equiv 3\cdot(2)^n + 4\times 2^n = 7\cdot 2^n\equiv 0\pmod 7$.

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$3^{2n+1}+2^{n+2}=3\times9^n+4\times2^n=7\times2^n+3\times(9^n-2^n)$

$=7\times2^n+3\times(9-2)(9^{n-1}+\cdots+2^{n-1})=\color{red}7\times2^n+3\times\color{red}7(9^{n-1}+\cdots+2^{n-1})$

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