Prove that the next integer to $(\sqrt3+1)^{2n}$ is $(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}$

Question

How the integer next to $(\displaystyle\sqrt3+1)^{2n}$ is $$(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}$$

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In my opinion, the next integer should be $$(\sqrt3+1)^{2n}-(\sqrt3-1)^{2n}+1$$

I saw an answer on math.stackexchange stating that the next integer after $(\sqrt3+1)^{2n}$ is the firstly written expression, but I cannot understand how$?$

Any help is greatly appreciated.

Answer 1

This is a follow-up to @RossMillikan's answer, but for general $n$. Applying the binomial expansion, $$(\sqrt{3}+ 1 )^{2n} = \sum_{k=0}^{2n} {2n\choose k} \sqrt{3}^{k} (1)^{2n-k}= \sum_{k=0}^{2n} {2n\choose k} {3}^{k/2} $$

Splitting the range of $k$ into even ($2j$) and odd ($2j+1$), $$(\sqrt{3}+ 1 )^{2n} = \sum_{j=0}^{n} {2n\choose 2j} {3}^{j} + \sum_{j=0}^{n-1} {2n\choose 2j+1} {3}^{j+\frac{1}{2}}$$

Following the same analysis for the other term, $$(\sqrt{3} - 1 )^{2n} = \sum_{k=0}^{2n} {2n\choose k} \sqrt{3}^{k} (-1)^{2n-k}= \sum_{j=0}^{n} {2n\choose 2j} {3}^{j} - \sum_{j=0}^{n-1} {2n\choose 2j+1} {3}^{j+\frac{1}{2}} $$

Then the sum is an integer, $$(\sqrt{3}+ 1 )^{2n} + (\sqrt{3} - 1 )^{2n} = 2\sum_{j=0}^{n} {2n\choose 2j} {3}^{j}$$

Let $a=(\sqrt{3}+ 1 )^{2n}, b = (\sqrt{3}- 1 )^{2n}, c = a + b = (\sqrt{3}+ 1 )^{2n} + (\sqrt{3} - 1 )^{2n}$. Since $b < 1$, then the gap between $c$ and $a$ is less than 1. Since we showed $c$ is already an integer, there cannot exist another integer between $a$ and $c$, since the gap $< 1$. Therefore, $c$ is the closest integer after $a$.

Answer 2

Your expression is not even an integer. If we take $n=1$ we have $$(\sqrt 3+1)^2=3+2\sqrt 3 +1\\ (\sqrt 3-1)^2=3-2\sqrt 3 +1\\ (\sqrt 3+1)^2-(\sqrt 3-1)^2+1=4\sqrt 3 +1$$ The recommended form has all the $\sqrt 3$s cancel and results in an integer. Then we note that $\sqrt 3-1 \lt 1$ so a power of it will be smaller yet.