$(2-\sqrt 3)^{2r-1}=1+m-n\sqrt 3$, therefore \[m=\frac{(2+\sqrt 3)^{2r-1}-2+(2-\sqrt 3)^{2r-1}}{2}=(\frac{(1+\sqrt 3)^{2r-1}+(1-\sqrt 3)^{2r-1}}{2^r})^2.\]

Nice problem, the problem before this in a problem set motivated my solution, I'll post both problems and solutions below. Quote: Find $a_{2020}$ for the sequence $(a_n)_{n \geq 1}$ defined by $a_1=1$ and \[a_{n+1}=2a_n+\sqrt{3a_n^2-2}, \qquad n=1,2, \ldots.\] Solution: Employ the fact $3=2^2-1.$ The recursion becomes $a_{n+1}=2a_n+\sqrt{(2^2-1)a_n^2-2}.$ Rearranging, we have $a_{n+1}-2a_n=\sqrt{(2^2-1)a_n^2-2}.$ Squaring both sides, we obtain \[(a_{n+1}-2a_n)^2=(2^2-1)a_n^2-2=a_{n+1}^2+2^2a_{n}^2-2a_{n+1}2a_n.\]Thus, $a_{n+1}^2+2^2a_{n}^2-2a_{n+1}2a_n=(2^2-1)a_n^2-2=2^2a_n^2-a_n^2-2.$ Simplifying, we obtain \[a_{n+1}^2-2a_{n+1}2a_n=-a_n^2-2 \implies a_{n+1}^2+a_n^2=2a_{n+1}2a_n-2.\]We know by definition of recursion then that $a_n^2+a_{n-1}^2=2a_n2a_{n-1}-2.$ Simplifying both of these equations, we obtain $a_{n+1}^2+a_n^2=4a_{n+1}a_n-2$ and $a_n^2+a_{n-1}^2=4a_na_{n-1}-2.$ Subtracting the two equations produces $a_{n+1}^2-a_{n-1}^2=4a_n(a_{n+1}-a_{n-1}).$ By difference of squares, we see $(a_{n+1}+a_{n-1})(a_{n+1}-a_{n-1})=4a_n(a_{n+1}-a_{n-1}),$ and it readily follows that $a_{n+1}+a_{n-1}=4a_n \implies a_{n+1}=4a_n-a_{n-1}.$ Now, notice that $a_1=1$ and $a_2=3.$ It suffices to find $a_{2020}$ in the recursion $a_1=1, a_2=3, a_n=4a_{n-1}-a_{n-2}.$ The characteristic equation of our recursion is $c^n=4c^{n-1}-c^{n-2} \implies c^2=4c-1.$ Solving the quadratic gives $c=2 \pm \sqrt{3}.$ Therefore, we know $a_n=\lambda_1(2+\sqrt{3})^n+\lambda_2(2-\sqrt{3})^n.$ Plugging in our values of $a_1$ and $a_2,$ we see that $1=\lambda_1(2+\sqrt{3})+\lambda_2(2-\sqrt{3})$ and $3=\lambda_1(2+\sqrt{3})^2+\lambda_2(2-\sqrt{3})^2.$ Solving this system of equations, we obtain $\lambda_1=\tfrac{3-\sqrt{3}}{6}$ and $\lambda_2=\tfrac{3+\sqrt{3}}{6},$ so \[a_n=\left(\frac{3-\sqrt{3}}{6}\right)(2+\sqrt{3})^n+\left(\frac{3+\sqrt{3}}{6}\right)(2-\sqrt{3})^n.\]Substituting in $n=2020,$ we get a very large integer, which makes this problem unsuitable to put in any competition. Also notice every integer in this sequence will be an integer, as every subsequent term is made up of a linear combination of the preceding two, and the sequence begins at $a_1=1, a_2=3.$ Now we move on to the actual problem! Quote: Prove that if $m,n,r$ are positive integers, and: \[1+m+n\sqrt{3}=(2+\sqrt{3})^{2r-1} \]then $m$ is a perfect square. We know by properties of radical conjugates that $1+m-n\sqrt{3} = (2-\sqrt{3})^{2r-1}.$ Along with the original expression, $1+m+n\sqrt{3}=(2+\sqrt{3})^{2r-1},$ add the two to obtain the equation $2+2m=(2-\sqrt{3})^{2r-1}+(2+\sqrt{3})^{2r-1},$ and dividing by $2,$ it follows \[m=\frac{1}{2}\left(\left(2-\sqrt{3}\right)^{2r-1}+\left(2+\sqrt{3}\right)^{2r-1}\right)-1.\]Recall in the previous problem where we had the recursion $a_n=4a_{n-1}-a_{n-2}.$ As the characterstic equation of this recursion has roots $2 \pm \sqrt{3},$ and we have $2\pm \sqrt{3}$ in our equation, it leads us to employing the recursion. Let $a_1=1, a_2=5, a_n=4a_{n-1}-a_{n-2}.$ Once again, we can write $a_n=\lambda_1(2+\sqrt{3})^n+\lambda_2(2-\sqrt{3})^n.$ We then have the system of equations $1=\lambda_1(2+\sqrt{3})+\lambda_2(2-\sqrt{3})$ and $5=\lambda_1(2+\sqrt{3})^2+\lambda_2(2-\sqrt{3})^2.$ Solving the system, we obtain $\lambda_1= \tfrac{\sqrt{3}-1}{2}$ and $\lambda_2=-\tfrac{\sqrt{3}+1}{2}.$ Thus we have the recursion \[\frac{\sqrt{3}-1}{2}\left(2+\sqrt{3}\right)^n - \frac{\sqrt{3}+1}{2}\left(2-\sqrt{3}\right)^n.\]Rewrite our fractions as $\tfrac{\sqrt{3}-1}{2}=\tfrac{1}{\sqrt{3}+1}$ and $-\tfrac{\sqrt{3}-1}{2}=\tfrac{1}{1-\sqrt{3}}.$ Consider some $a_r^2.$ Then \begin{align*} a_r^2 &= \left(\frac{1}{\sqrt{3}+1}\left(2+\sqrt{3}\right)^r + \frac{1}{1-\sqrt{3}}\left(2-\sqrt{3}\right)^r\right)^2 \\ &= \frac{(2+\sqrt{3})^{2r}}{(\sqrt{3}+1)^2} + \frac{(2-\sqrt{3})^{2r}}{(1-\sqrt{3})^2} - \frac{2((2-\sqrt{3})(2+\sqrt{3}))^r}{(1-\sqrt{3})(1+\sqrt{3})} \\ &= \frac{(2+\sqrt{3})^{2r}}{(\sqrt{3}+1)^2} + \frac{(2-\sqrt{3})^{2r}}{(1-\sqrt{3})^2} -1 \\ &= \frac{1}{2}\left(\left(2-\sqrt{3}\right)^{2r-1}+\left(2+\sqrt{3}\right)^{2r-1}\right)-1 \\ &=m. \end{align*}As in the previous problem, because every subsequent term is made up of a linear combination of the preceding two, and the sequence begins with $a_1=1, a_2=5,$ it follows all $a_k$ are integers, and thus $m$ is a perfect square, as required.