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 $n$?

  • $\begingroup$ What have you tried? If we write out the first few values of $\sqrt {a_n-1}$ for odd $n$ we get $\{1,5,19,71,265,989,\cdots\}$ which appears to be A001834. Perhaps one can prove that equality. $\endgroup$
    – lulu
    Commented Aug 1, 2017 at 14:56

2 Answers 2


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,


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.


A first step:

Define integer sequences $x_n$ and $y_n$ by $$ (2+\sqrt{3})^{2n+1} = x_n+y_n\sqrt{3}. $$ Then $x_0=2$ and $y_0=1$, and furthermore you get the recursion \begin{eqnarray*} x_{n+1} &=& 7x_n+12y_n\\ y_{n+1} &=& 4x_n+7y_n \end{eqnarray*}. The first equation gives $12y_n=x_{n+1}-7x_n$, and plugging this into the second equation gives $$ y_{n+1} = 4x_n+7y_n = 4x_n+\frac{7}{12}(x_{n+1}-7x_n) = \frac{1}{12}(7x_{n+1}-x_n).$$ This yields $12y_n=7x_n-x_{n-1}$, and plugging it into the first recursion yields $$ x_{n+1} = 14x_n-x_{n-1}. $$ Together with $x_0=2$ and $x_1=26$, this nicely describes the sequence of odd-indexed numbers $a_{2n+1}=x_n$. Hence we want to show that every $x_n-1$ is a square.


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