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Prove that $$3^{2n+3}+2^{n+3}$$ is divisible with $7$ for every $n \in \Bbb N$

$$3^{2n}\cdot27+2^{n}\cdot8=3^{2n}(28-1)+2^{n}(7+1)=\left(3^{2n}\cdot28+2^{n}\cdot7\right)-\left(9^{n}-2^{n}\right)=$$

$$\left(3^{2n}\cdot28+2^{n}\cdot7\right)-(9-2)\left(9^{n-1}+\ldots+2^{n-1}\right)$$ if $n=2k+1$. If $n=2k$ is also true. So the number is divisible with $7$.

Is there any proof without using $a^{n}-b^{n}$

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without $\mod$ most elementary proof. –  Iuli Feb 23 '13 at 23:19

2 Answers 2

up vote 5 down vote accepted

I'll leave to you to show for $\,n=1\,$. Assume now the claim's true for $\,n\,$ and we show for $\,n+1\,$:

$$3^{2(n+1)+3}+2^{n+1+3}=9\cdot 3^{2n+3}+2\cdot2^{n+3}=2\left(3^{2n+3}+2^{n+3}\right)+7\cdot3^{2n+3}=\ldots $$

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Hint $\rm\ mod\ 7\!:\,\ 3^{2n+3}\!+2^{n+3}\equiv\, 27\cdot 9^n\! + 8\cdot 2^n \equiv\, 35\cdot 2^n\equiv 0\ \ $ by $\rm\ 9\equiv 2;\ 35\equiv 0$

Remark $\ $ Per your comment, for a proof without mod, subtract and add $\rm\,27\cdot 2^n\,$ above, i.e.

$\rm\quad 27\cdot 9^n\! + 8\cdot 2^n =\, 27\, (9^n\!-2^n)\! + (\color{#C00}{27\!+\!8})\cdot 2^n $ is divisible by $\,7\,$ by $\rm\ 9\!-\!2\,|\,9^n\!-2^n,\,$ and $\rm\, 7\,|\, \color{#C00}{35}$

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+1: Modular arithmetic really makes many things simple once you're acclimated to it! –  Hurkyl Feb 23 '13 at 23:33

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