USA IMO

First Solution. The equation is symmetric in $m$ and $n$. The solutions are the unordered pairs $$\{5F_{2k}^2,5F_{2k+2}^2\},\{L_{2k-1}^2,L_{2k+1}^2\},$$ where $k$ is a nonnegative integer and $\{F_j\}$, $\{L_j\}$ are the Fibonacci and Lucas sequences, respectively — that is, the sequences defined by $F_1=F_2=1$, $L_1=1$, $L_2=3$, and the recursive relations $F_{j+2}=F_{j+1}+F_j$ and $L_{j+2}=L_{j+1}+L_j$ for $j\ge 1$. Note that we amended the Lucas sequence by considering $L_1=-1$ and $L_0=2$. Let $g=\gcd (m,n)$ and write $m=g{m}_1$ and $n=g{n}_1$. Because $9mn$ is a perfect square, $m_1$ and $n_1$ are perfect squares. Let $m_1=x^2$ and $n_1=y^2$. The given condition becomes $$(g{x}^2+g{y}^2-5{)}^2=9g^2{x}^2{y}^2.$$ Taking the square root on both sides yields $$g(x^2+y^2)-5=\pm 3gxy,$$ or $$g(x^2+y^2\pm 3xy)=5.$$ If $g(x^2+y^2+3xy)=5$, then $x^2+y^2+3xy\le 5$, implying that $x=y=g=1$ and $(m,n)=(1,1)$. Otherwise, $g(x^2+y^2-3xy)=5$ and $g=1$ or $5$. Fix $g$ equal to one of these values, so that $$x^2-3xy+y^2=\frac{5}{g}.(1)$$ Let $\{a,b\}$ denote an unordered pair of numbers. We call $\{a,b\}$ a g-pair if $\{x,y\}=\{a,b\}$ satisfies (1) and a and b are positive integers. Also, we call $\{p,q\}$ smaller (respectively, larger) than $\{r,s\}$ if $p+q$ is smaller (respectively, larger) than $r+s$. Suppose that $\{a,b\}$ is a g-pair. View (1) as a monic quadratic in $x$ with $y=b$ constant. The coefficient of $x$ in a monic quadratic function $(x-r_1)(x-r_2)$ equals $-(r_1+r_2)$, implying that $\{3b-a,b\}$ should also satisfy (1). Indeed, $$b^2-3b(3b-a)+(3b-a{)}^2=a^2-3ab+b^2=\frac{5}{g}.$$ Also, if $b>2$, note that $$a^2-3ab+b^2=\frac{5}{g}<b^2.$$ It follows that $a^2-3ab<0$ and so $3b-a>0$. Thus, if $\{a,b\}$ is a $g$-pair with $b>2$, then $\{b,3b-a\}$ is a $g$-pair as well. Also note that for $a^{\prime }=b$ and $b^{\prime }=3b-a$, $\{a^{\prime },3b^{\prime }-a^{\prime }\}=\{a,b\}$. Furthermore, if $a\ge b$, note that $a\ne b$ because otherwise $-a^2=5/g>0$, which is impossible. Thus, $a>b$ and $$a^2-3ab+b^2=\frac{5}{g}>b^2-a^2,$$ which implies that $a(2a-3b)>0$ and hence $a+b>b+(3b-a)$ and also $3b-a>b$. Thus, $\{a^{\prime },b^{\prime }\}$ is a smaller $g$-pair than $(a,b)$ with $a^{\prime }\ge b^{\prime }$. Given any $g$-pair $\{a,b\}$ with $b\le a$, if $b\le 2$ then $a$ must equal $r(g)$, where $r(5)=3$ if $r(1)=4$. Otherwise, according to the above observation we can repeatedly reduce it to a smaller $g$-pair until $\min (a,b)\le 2$ — that is, to the $g$-pair $(r(g),1)$. Beginning with $\{r(g),1\}$, we reverse the reducing process so that $\{x,y\}$ is replaced by the larger $g$-pair $\{3x-y,x\}$. Moreover, this must generate all $g$-pairs since all $g$-pairs can be reduced to $\{r(g),1\}$. We may express these possible pairs in terms of the Fibonacci and Lucas numbers; for $g=1$, observe that $L_2=1$, $L_3=4=r(1)$, and that $$\begin{array}{rl}{L}_{2k+3}& =L_{2k+2}+L_{2k+1}=(L_{2k+1}+L_{2k})+L_{2k+1}\\ & =(L_{2k+1}+(L_{2k+1}-L_{2k-1}))+L_{2k+1}\\ & =3L_{2k+1}-L_{2k-1}\end{array}$$ for $k\ge 0$. For $g=5$, the Fibonacci numbers satisfy an analogous recursive relation, and $F_2=1$, $F_4=3=r(5)$. Therefore, $\{m,n\}=\{L_{2k-1}^2,L_{2k+1}^2\}$ and $\{m,n\}=\{5F_{2k}^2,5F_{2k+2}^2\}$ for $k\ge 0$.

---

Alternative solution.

Second Solution. As we have seen in the first solutions, it suffices to solve the equations $$x^2+y^2-3xy=1(2)$$ and $$x^2+y^2-3xy=5.(3)$$ The first equation is equivalent to $(2x-3y{)}^2-5y^2=4$, which is a Pell's equation of the form $$u^2-d{v}^2=4.(4)$$ All the solutions $(u,v)=(u_k,v_k)$, $k\ge 1$, of equation (4) are given by $$\frac{1}{2}(u_k+v_k\sqrt{d})={\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k,k\ge 1,(5)$$ where $(u,v)=(u_1,v_1)$ is the solution to (4) with least positive $u$ value. From (5) it is not difficult to see that $$u_k={\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k+{\left(\frac{u_1-v_1\sqrt{d}}{2}\right)}^k$$ and $$v_k=\frac{1}{\sqrt{d}}\left[{\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k-{\left(\frac{u_1-v_1\sqrt{d}}{2}\right)}^k\right]$$ for all $k\ge 1$. In our case, $(u_1,v_1)=(3,1)$. Hence $$\begin{array}{rl}{y}_k& =v_k=\frac{1}{\sqrt{5}}\left[{\left(\frac{3+\sqrt{5}}{2}\right)}^k-{\left(\frac{3-\sqrt{5}}{2}\right)}^k\right]\\ & =\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{2k}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{2k}\right]=F_{2k},\end{array}$$ where $\{F_n{\}}_n^{\infty }$ is the Fibonacci sequence, and $$2x_k-3y_k=u_k={\left(\frac{3+\sqrt{5}}{2}\right)}^k-{\left(\frac{3-\sqrt{5}}{2}\right)}^k,$$ implying $$\begin{array}{rl}{x}_k& =\frac{1}{2}(u_k+3v_k)\\ & =\frac{1}{2}\left[\left(\frac{3}{\sqrt{5}}+1\right){\left(\frac{3+\sqrt{5}}{2}\right)}^k-\left(\frac{3}{\sqrt{5}}-1\right){\left(\frac{3-\sqrt{5}}{2}\right)}^k\right]\\ & =\frac{1}{\sqrt{5}}\left[{\left(\frac{3+\sqrt{5}}{2}\right)}^{k+1}-{\left(\frac{3-\sqrt{5}}{2}\right)}^{k+1}\right]\\ & =\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{2k+2}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{2k+2}\right]=F_{2k+2}.\end{array}$$ It follows that $m_k=5x_k^2=5F_{2k+2}^2$ and $n_k=5y_k^2=5F_{2k}^2$, for $k\ge 1$.

Equation (3) is equivalent to $(2x-3y{)}^2-5y^2=20$. Setting $2x-3y=5v$ and $y=u$, the equation reduces to $u^2-5v^2=-4$. This is a Pell's equation of the form $$u^2-d{v}^2=-4,(6)$$ whose solutions $(u,v)=(u_k,v_k)$, $k=1,3,5,\dots$, are given by $$\frac{1}{2}(u_k+v_k\sqrt{d})={\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k,k=1,3,5,\dots ,$$ where $(u,v)=(u_1,v_1)$ is the solution to (6) with least positive $u$ value. It follows that $$u_k={\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k+{\left(\frac{u_1-v_1\sqrt{d}}{2}\right)}^k$$ and $$v_k=\frac{1}{\sqrt{d}}\left[{\left(\frac{u_1+v_1\sqrt{d}}{2}\right)}^k-{\left(\frac{u_1-v_1\sqrt{d}}{2}\right)}^k\right]$$ for $k=1,3,5,\dots$. In our case, $(u_1,v_1)=(1,1)$. Hence $$u_k={\left(\frac{1+\sqrt{5}}{2}\right)}^k+{\left(\frac{1-\sqrt{5}}{2}\right)}^k=L_k,$$ where $\{L_n{\}}_{n=1}^{\infty }$ is the Lucas sequence, and $$v_k=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^k-{\left(\frac{1-\sqrt{5}}{2}\right)}^k\right]=F_k,$$ for $k=1,3,5,\dots$. Hence $y_k=L_k$, $k=1,3,5,\dots$, and $$\begin{array}{rl}{x}_k& =\frac{1}{2}(3L_k+5F_k)\\ & =\frac{1}{2}\left[(3+\sqrt{5}){\left(\frac{1+\sqrt{5}}{2}\right)}^k+(3-\sqrt{5}){\left(\frac{1-\sqrt{5}}{2}\right)}^k\right]\\ & ={\left(\frac{1+\sqrt{5}}{2}\right)}^{k+2}+{\left(\frac{1-\sqrt{5}}{2}\right)}^{k+2}\\ & =L_{k+2}.\end{array}$$ Therefore $m_k=5L_{k+2}^2$, $n_k=5L_k^2$, $k=-1,1,3,5,\dots$. Here we extended the definition of Lucas number to $L_0=2$ and $L_{-1}=-1$.