Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the integer just above 7.3 . It is 8. Then , Q wants you to prove that 8 is divisible by $2^{n+1}$.
\begin{equation}
\begin{array}{l}
(\sqrt{3}+1)^{2 n}=(4+2 \sqrt{3})^{n}=2^{n}(2+\sqrt{3})^{n}\\
=2^{n}(2+\sqrt{3})^{n}=2^{n}\left[^n C _{0}2^{n}+^n C_{1} 2^{n-1} \sqrt{3}+^n C_{{2}} 2^{n-2} \sqrt{3}^{2}+\right.\\
\begin{array}{l}
(\sqrt{3}-1)^{2 n}=(4-2 \sqrt{3})^{n}=2^{n}(2-\sqrt{3})^{n} \\
=2^{n}(2-\sqrt{3})^{n}=2^{n}\left[{ }^{n} c_{0} 2^{n}-n_{C}, 2^{n-1} \sqrt{3}+{ }^{n} c_{2} 2^{n-1} \sqrt{1}^{2} \ldots\right.
\end{array}\\
I+f+f)=2^{n}\left[2(\text { Integer) }]=2^{n+1}\right. \text { . Integer }\\
I+1=2^{n+1} \text { . Integer }
\end{array}
\end{equation}
In the image is the way this question is solved.
My Q from this method of solving is that if we notice at the end , we somehow got $ 2^{n+1}$. If the Q has taken some other value like $3^{n+3}$ or Something else. Then , it was not possible to prove this question.
What is another method to prove this Q or can you help me justify that the above method can be used for all kinds of Q.
Thank you.
 A: Note that $0\lt \sqrt 3 -1 \lt 1$
Now look at the numbers $a=1+\sqrt 3, b=1-\sqrt 3$ with $a+b=2, ab=-2$ which are roots of the quadratic $x^2-2x-2=0$
Then with $v_n=a^n+b^n$ we have $v_{n+2}-2v_{n+1}-2v_n=0$ (spot the coefficients) or $v_{n+2}=2(v_{n+1}+v_n)$ now $|b^n|\lt 1$ and indeed $0\lt b^{2n}\lt 1$ so the difference $v_{2m}-a^{2m}=b^{2m}$ is small and positive.
It will be found that $v_n$ is an integer close to $a^n$ for all $n$, and the sign of the difference for even indices is right, and you can run an induction to complete the proof.
If you understand this you will be able to do the same trick with  $A=a^2$ and $B=b^2$ to get a rather simpler induction.
For these kinds of questions with powers of an irrational number "miraculously close" to integers, there is very often a recurrence to be found lurking in the background.

If $a$ and $b$ are the roots of $x^2-px+q$=0 and we put $u_n=Ca^n+Db^n$ we can reason as follows:
$$a^2-pa+q=0$$ because $a$ is a root. Multiply through bay $a^n$ to give $$a^{n+2}-pa^{n+1}+qa^n=0$$
Now multiply through by the constant $C$ $$Ca^{n+2}-Cpa^{n+1}+Cqa^n=0$$
Similarly with the other root $b$ and the constant $D$ $$Db^{n+2}-Dpb^{n+1}+Dqb^n=0 $$
Now add these last equations to obtain $$Ca^{n+2}+Db^{n+2}-p\left(Ca^{n+1}+Db^{n+1}\right)+q\left(Ca^n+Db^n\right) = 0 = u_{n+2}-pu_{n+1}+qu_n$$
And this works for any constants $C, D$ and hence for $C=D=1$.

We can also argue that $p=a+b$ and $q=ab$ so that
$$u_{n+2}-pu_{n+1}+qu_n=$$$$=Ca^{n+2}+Db^{n+2}-(a+b)Ca^{n+1}-(a+b)Db^{n+1}+abCa^n+abDb^n=0$$ because the terms all cancel.
A: Part of the problem is that the image was a bit obscured and so the transcription was a bit off. Here is a more accurate transcription with tags added for reference:

$$\require{cancel}
\left(\sqrt3+1\right)^{2n}=\left(4+2\sqrt3\right)^n=2^n\left(2+\sqrt3\right)^n\tag1
$$
$$
\!\!\!\!I+f=2^n\left(2+\sqrt3\right)^n=2^n\left[{}^nC_02^n+\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2+\dots\right]\tag2
$$
$$
\left(\sqrt3-1\right)^{2n}=\left(4-2\sqrt3\right)^n=2^n\left(2-\sqrt3\right)^n\tag3
$$
$$
f'=2^n\left(2-\sqrt3\right)^n=2^n\left[{}^nC_02^n-\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2-\dots\right]\tag4
$$

$$
I+\bbox[3px,border:1px solid black]{f+f'}=2^n[2(\text{Integer})]=2^{n+1}\cdot\text{Integer}\tag5
$$
$$
I+1=2^{n+1}\cdot\text{Integer}\quad\checkmark\tag6
$$

$(1)$ is just simple algebraic manipulation
$(2)$ is expanding via the Binomial Theorem ($I$ and $f$ are the integer and fraction parts)
$(3)$ is $(1)$ with the substitution $\sqrt3\mapsto-\sqrt3$; note that $\left(\sqrt3-1\right)^{2n}=\left(-\sqrt3+1\right)^{2n}$
$(4)$ is expanding via the Binomial Theorem
$(5)$ adding $(2)$ and $(4)$ cancels all of the terms with $\sqrt3$ to an odd power
$\phantom{\text{(5)}}$ and doubles all of the terms with $\sqrt3$ to an even power (these terms are integers)
$(6)$ since $0\lt\left(4-2\sqrt3\right)\lt1$, $f'\in(0,1)$ and by definition, $f\in[0,1)$
$\phantom{\text{(6)}}$ since $I+f+f'\in\mathbb{Z}$, we must have $f+f'=1$

Here is, in my experience, a more usual proof of this.
$\left(1\pm\sqrt3\right)^2=4\pm2\sqrt3$ are roots of $x^2-8x+4$. Therefore, the solution to the linear recurrence equation
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[3pt]
&=4(2u_{n-1}-u_{n-2})\tag7
\end{align}
$$
is
$$
\begin{align}
u_n
&=a\left(1+\sqrt3\right)^{2n}+b\left(1-\sqrt3\right)^{2n}\\[3pt]
&=a\left(4+2\sqrt3\right)^n+b\left(4-2\sqrt3\right)^n\tag8
\end{align}
$$
In particular, the sequence with $a=b=1$ starts out with
$$
u_0=2,u_1=8,u_2=56\tag9
$$
Note that $\left.2^{n+1}\,\middle|\,u_n\right.$ for $n=1$ and $n=2$. Then $(7)$ guarantees that
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[6pt]
&=8\cdot2^n\frac{u_{n-1}}{2^n}-4\cdot2^{n-1}\frac{u_{n-2}}{2^{n-1}}\\
&=2^{n+1}\left(4\frac{u_{n-1}}{2^n}-\frac{u_{n-2}}{2^{n-1}}\right)\tag{10}
\end{align}
$$
By induction, $(9)$ and $(10)$ show that $\left.2^{n+1}\,\middle|\,u_n\right.$ for all $n\ge1$.
Since $u_n\in\mathbb{Z}$,
$$
\begin{align}
u_n
&=\left(1+\sqrt3\right)^{2n}+\overbrace{\left(1-\sqrt3\right)^{2n}}^{\text{in }(0,1)}\\[3pt]
&=\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\tag{11}
\end{align}
$$
Therefore, $(10)$ and $(11)$ show that
$$
\left.2^{n+1}\,\middle|\,\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\right.\tag{12}
$$
