<s>[INST] Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........? Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?
I am not well versed with binomial theorem hence I am not able to proceed. I am also interested in the different ways this problem can be solved.
Any help would be appreciated.
 [/INST] From binomial expansion it follows that
$$[(2+\sqrt{3})^n+(2-\sqrt{3})^n]$$ $$=[2^n+{n \choose 2} 2^{n-2} 3+{n \choose 4}2^{n-4} 3^2+...]=2K, K \in N$$
And $$[(2+\sqrt{3})^n-(2-\sqrt{3})^n]$$ $$=2\sqrt{3}[2^{n-1}+{n \choose 1} 2^{n-1}+{n \choose 3}2^{n-3} 3+...]=2\sqrt{3}L, L \in N$$
$$S_1=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}=2^n[(2+\sqrt{3})^n+(2-\sqrt{3})^n]=2^{n+1} K.$$
$$S_2=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}=S_1+2^n \sqrt{3}[(2+\sqrt{3})^{n}-(2-\sqrt{3})^n]=3.2^{n+1}L$$
Hence, both $S_1$ and $S_2$ are divisible by $2^{n+1}$.
By binomial expansio
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