Who is Vieta?

Using the binomial expansion, $(\sqrt3+\sqrt2)^n+(\sqrt3-\sqrt2)^n$ $= \left[ \sqrt3^n + \begin{pmatrix} n \\ 1 \end{pmatrix} \sqrt3^{n-1}\sqrt2 + \begin{pmatrix} n \\ 2 \end{pmatrix} \sqrt3^{n-2}\sqrt2^2+ \cdots + \sqrt2^n\right] + \left[ \sqrt3^n - \begin{pmatrix} n \\ 1 \end{pmatrix} \sqrt3^{n-1}\sqrt2 + \begin{pmatrix} n \\ 2 \end{pmatrix} \sqrt3^{n-2}\sqrt2^2+ \cdots + \sqrt2^n\right]$ $= 2 \left[ \sqrt3^n + \begin{pmatrix} n \\ 2 \end{pmatrix} \sqrt3^{n-2}\sqrt2^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} \sqrt3^{n-4}\sqrt2^4 + \cdots + \sqrt2^n \right]$ Since $n$ is an even positive integer, we can write $n=2a$ for some integer $a$ : $= 2 \left[ \sqrt3^{2a} + \begin{pmatrix} n \\ 2 \end{pmatrix} \sqrt3^{2a-2}\sqrt2^2 + \begin{pmatrix} n \\ 4 \end{pmatrix} \sqrt3^{2a-4}\sqrt2^4 + \cdots + \sqrt2^{2a} \right]$ $= 2 \left[ 3^{a} + \begin{pmatrix} n \\ 2 \end{pmatrix} 3^{a-1} 2 + \begin{pmatrix} n \\ 4 \end{pmatrix} 3^{a-2}2^2 + \cdots + 2^a \right]$ which is an integer. Therefore the statement is $\large \color{#20A900}{\boxed{\text{True}}}$ .