# A question in number theory

Prove that for all $n \in \mathbb N$ the quotient of $$\frac{3^{2^n}}{2^n}$$ is an even number. Maybe we have to check a more general case, namely $\displaystyle\frac{3^k}{k}$. What would you do here? Thanks!

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You mean that $\lfloor 3^{2^n}/2^n\rfloor$ is an even integer, right? – Pedro Tamaroff Sep 28 '12 at 18:13
@Peter Tamaroff: I just wrote it the way it's written in my textbook. Sure. – user 1618033 Sep 28 '12 at 18:14
Try to write that number as $(3^k+1)/2^n$ – Pedro Tamaroff Sep 28 '12 at 18:15
Now that there are two answers - may I assume that the real question was to show that $\frac{3^{2^n}-1}{2^n}$ is even? – Hagen von Eitzen Sep 28 '12 at 18:25

Hint: $$3^{2^n}-1^{2^n}=(3^{2^{n-1}}+1^{2^{n-1}})(3^{2^{n-1}}-1^{2^{n-1}}).$$ Continue.

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nice and simple :-) Thanks! – user 1618033 Sep 28 '12 at 18:18

From Euler's theorem, $3^{2^{n-1}} \equiv 1 \pmod {2^n}$

Hence, $$3^{2^{n-1}} = M2^n + 1 \implies 3^{2^n} = \left(3^{2^{n-1}} \right)^2 = \left(M2^n + 1 \right)^2 = 2^n \left( M^2 2^n + 2M\right) + 1$$

Hence, $$\dfrac{3^{2^n}}{2^n} = 2 M \left(2^{n-1} M + 1 \right) + \dfrac1{2^n}$$

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Thanks for your solution. (+1) – user 1618033 Sep 28 '12 at 18:23