Show that $ \left\lceil( \sqrt3 +1)^{2n}\right\rceil$ where $n \in \mathbb{N}$ is divisible by $2^{n+1}$.

I wrote the binomial expansion of $ ( \sqrt3 +1)^{2n}$ and $( \sqrt3 -1)^{2n}$ and then added them to confirm that the next integer is even.
Afterwards I applied $AM \ge GM$ on the two terms to get $ ( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} \ge (2^{n+1})$.

Now I'm unable to figure out the next step. Any help would be appreciated. :)

  • $\begingroup$ No induction pls $\endgroup$ – Harshal Gajjar Jan 21 '15 at 6:05
  • $\begingroup$ A good start, but why the wish to avoid induction? To me it feels like the most natural approach. I would do a two step induction based on the recurrence relation. $\endgroup$ – Jyrki Lahtonen Jan 21 '15 at 6:11
  • 1
    $\begingroup$ OK, no induction, then no solution. $\endgroup$ – André Nicolas Jan 21 '15 at 6:12
  • $\begingroup$ Ah.. I found this question in an objective paper with RHS as one of the options. This makes me feel that there might be a better method of solving this question. :) $\endgroup$ – Harshal Gajjar Jan 21 '15 at 6:13
  • $\begingroup$ @AndréNicolas Is there really no other method of solving? $\endgroup$ – Harshal Gajjar Jan 21 '15 at 6:14

$$(\sqrt{3}\pm 1)^2=(4 \pm 2\sqrt{3})=2(2 \pm \sqrt{3})$$

Therefore $$( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} =2^n \left[(2 + \sqrt{3})^n+(2 - \sqrt{3})^n \right] $$

Now, use the Binomial Theorem to prove that $(2 + \sqrt{3})^n+(2 - \sqrt{3})^n $ is an even integer.


Let: $$ A_n = (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n} = (4+2\sqrt{3})^n+(4-2\sqrt{3})^{n}.\tag{1} $$ Since $0<4-2\sqrt{3}<\frac{2}{3}$, we have that $A_n$ is the integer giving the ceiling of $(\sqrt{3}+1)^{2n}$.

Now we have: $$ A_0 = 2, \quad A_1 = 8,\qquad A_{n+2} = 8 A_{n+1} - 4 A_{n}\tag{2} $$ and if we take: $$ \nu_2(n) = \max\{m\in\mathbb{N}: 2^m\mid n\} \tag{3}$$ it happens that: $$ \nu_2(A_n)\geq n+1\tag{4} $$ can be proved by induction from $(2)$. We can also factor a $2^n$ from the RHS of $(1)$ then just study the parity of the sequence given by: $$ B_n = (2+\sqrt{3})^n+(2-\sqrt{3})^n,\tag{5}$$ for which: $$ B_0=2,\quad B_1=4,\quad B_2=14,\quad B_{n+2}=4B_{n+1}-B_n\equiv -B_n\pmod{4}.\tag{6}$$ That gives:

$$ \nu_2(A_n) = \left\{\begin{array}{rl}n+1 & \text{if }n\text{ is odd,}\\n+2 &\text{if }n\text{ is even.}\end{array}\right.\tag{7}$$


Let $R=(\sqrt{3}+1)^{2n}=I+f$ where $I$ is the integeral part of $R$ and $f$ is the fractional part of $R$.
Let $r=(\sqrt{3}-1)^{2n}$
Now, we know that $r<1$ as $\sqrt{3}-1$ is less than $1$ and $n\geq 1$. Also,
$R+r=2({2n\choose0}\sqrt{3}^{0}+{2n\choose2}\sqrt{3}^{2}+\ldots+{2n\choose2n}\sqrt{3}^{2n})=2k$ where $k$ is a positive integer. Since, $r$ was less than $1$, this implies that $r+f=1$ and $I=2k-1$.
This means that $I$ is odd so $I+1$ is even.
EDIT: I am not deleting this answer so others can see how OP reached at the conclusion that $\lceil (\sqrt{3}+1)^{2n}\rceil$ is even.

  • $\begingroup$ I think the question was about the next larger integer (rounded up). $\endgroup$ – Martin R Jan 21 '15 at 8:31
  • $\begingroup$ But the greatest integer function returns the rounded down value for positive numbers. OP asked to show that the greatest integer of $(\sqrt{3}+1)^{2n}$ is divisble by $2^{n+1}$, which is false. $\endgroup$ – AvZ Jan 21 '15 at 8:46
  • $\begingroup$ You calculated $\lfloor x \rfloor$, but $\lceil x \rceil$ denotes the smallest integer larger than $x$, e.g. $\lceil 2.1 \rceil = 3$. Originally the question explicitly asked about "the integer immediately greater than ...". This was then changed to use the symbolic notation $\lceil \ldots \rceil$. $\endgroup$ – Martin R Jan 21 '15 at 8:55
  • $\begingroup$ Oh, my bad. I thought he wrote the greatest integer function $[x]$ $\endgroup$ – AvZ Jan 21 '15 at 8:59

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