## zmudz one year ago Let $$n$$ be a positive integer. Show that the smallest integer greater than $$(\sqrt{3} + 1)^{2n}$$is divisible by $$2^{n+1}.$$ Hint: Prove that $$\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{2n} + (\sqrt{3}-1)^{2n}.$$

1. amilapsn

you can use mathematical induction with binomial expansion.

2. anonymous

Id probably say make a recurrence relation

3. zzr0ck3r

A number can't be equal to itself plus a nonzero number...

4. anonymous

Isn't that$\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{n} + (\sqrt{3}-1)^{n}.$

5. amilapsn

$\exists ! 0.abcd.... ~s.t.$ $\lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+0.abcd.....$ $But~ 0<(\sqrt{3}-1)^{2n}< 1\forall n\in \mathbb{Z}^+$ By binomial expansion: $\small (\sqrt{3}+1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots+^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)$ $\small (\sqrt{3}-1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}-^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots-^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)$ $\small \therefore (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}=2 \left[^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_{2} (\sqrt{3})^{2n-2}+\ldots+^{2n}C_{2n-2} (\sqrt{3})^{2}+^{2n}C_{2n} 1\right]$ Since all the terms have even exponents of $$\sqrt{3}$$ this will be an integer... $\therefore \lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ By mathematical induction we can show that.. $2^{n+1}|\lceil{(\sqrt{3}+1)^{2n}}\rceil$

6. zzr0ck3r

suppose $$a=a+b$$ where $$b\ne 0$$. Then by cancelation $$b=0$$ a contradiction. :)

7. amilapsn

@zzr0ck3r you've forgotten the ceiling... :) for example $$\lceil 2.5\rceil =3$$

8. zzr0ck3r

Ahh, sorry on my phone and I thought it was just $$[$$. lol. I was going nuts...