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I am not able to solve this problem

Show that the integer just greater than $( \sqrt{3} + 1 )^{2m}$ contains $2^{m+1}$ as a factor.

Any help would be thoroughly appreciated

coffeemath
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user34304
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  • user34304-- just edited it to be in math format. I checked it and it looks like this is what it should be (true for first several values of $m$ in this form). Let me know if it isn't right now... – coffeemath Jan 26 '14 at 08:48

2 Answers2

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This is a Pisot numbers issue.

Let $u_n=(4+2\sqrt{3})^n+(4-2\sqrt{3})^n$. Then we have the recurrence relation $u_{n+2}=8u_{n+1}-4u_n$, and $u_0=2,u_1=8$.

It is easy then to see by induction on $n$ that $u_n$ is an integer multiple of $2^{n+1}$ for every $n$.

Further, $0 \leq (4-2\sqrt{3})^n \leq \frac{1}{2}$, so that $\lceil (4+2\sqrt{3})^n \rceil$ is in fact $u_n$, and we are done.

Ewan Delanoy
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You can write the expression as $\lceil (\sqrt 3+1)^2 \rceil ^m = (\sqrt 3^2 +2\sqrt 3+1^2 )^m=(4+2 \sqrt 3)^m$ and then you can extrac a factor $2$ out of the $m$ power: $(4+2\sqrt 3)^m=2^m(2+\sqrt 3)^m$. Now you have a $2^m$ factor, and you only have to see that the first integer that is bigger than $(2+\sqrt 3)^m$ is always even because the first digit of $(2+\sqrt 3)^m$ is always odd. So, the integer bigger than $(2+\sqrt 3)^m$, $T$, is of the form $T=2q$ and $2^m T=2^m 2q = 2^{m+1}q$.

Karolis Juodelė
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M159
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    There is a lack of an argument for $\lceil (2+\sqrt 3)^m \rceil$ being even. – Karolis Juodelė Jan 26 '14 at 08:59
  • Yes, but I don't have any more ideas :). – M159 Jan 26 '14 at 09:11
  • From looking a the numbers, it appears that $(2+\sqrt 3)^m$ not only preserves the parity of its integer part, but also approaches $1$ in its fractional part. – Karolis Juodelė Jan 26 '14 at 10:50
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    If $c(x)$ means ceiling of $x$, note one cannot extract the power of $2$ from inside, without justification. It isn't automatic. If $\alpha$ is irrational, one cannot say in general that $$c[(2a)^n]=c[2^n\alpha^n]=2^nc[\alpha^n].$$ – coffeemath Jan 26 '14 at 11:15