Prove that $\sqrt{m-1}$ is an integer 
Let $k$ be a fixed odd positive integer. Let $m,n$ be positive integers such that $(2+\sqrt{3})^k=m+n\sqrt{3}$. Prove that $\sqrt{m-1}$ is an integer.

Let $(2+\sqrt{3})^n = a_n+b_n\sqrt{3}$. From $$a_{n+1}+b_{n+1}\sqrt{3} = (a_n+b_n\sqrt{3})(2+\sqrt{3}) = (2a_n+3b_n)+(a_n+2b_n)\sqrt{3},$$ we get $a_{n+1} = 2a_n+3b_n$ and $b_{n+1} = a_n+2b_n$ with $a_1 = 2$ and $b_1 = 1$.
How can we show that $\sqrt{a_n-1}$ is an integer for odd $n$?

Since $k$ is odd $m-1$ is not multiple of $3$. This is because it is a sum of multiples of $3$ and a $2^{k}=2\cdot 4^{(k-1)/2}=2\mod{3}$.
Now, $m^2-3n^2=1$. So,
$$(m-1)(m+1)=3n^2$$
Since $m-1$ is not divisible by $3$, the $3$ divides the factor $m+1$. 
The only common divisor of $m-1$ and $m+1$ can be $2=(m+1)-(m-1)$. But $m$ is even. Therefore, $m-1$ and $m+1$ are relatively prime.
So, $m-1$ is a square.
