China Team Selection Test

$$((k-2{)}^2+1)xy=(x+y-k{)}^2+1,\stackrel{◯}{1}$$ we prove ① has infinitely many positive odd solutions $(x,y)$. Obviously $(1,1)$ is one positive odd solution, let $(x_1,y_1)=(1,1)$. Assume $(x_i,y_i)$ is one positive odd solution of ①, and $x_i\le y_i$, let $x_{i+1}=y_i$, $y_{i+1}=(k-1)(k-3)y_i+2k-x_i$. Since ① can be written as $$x^2-((k-1)(k-3)y+2k)x+(y-k{)}^2+1=0,$$ by Vieta's theorem $(x_{i+1},y_{i+1})$ is also one integer solution of ①. Since $x_i,y_i$ and $k$ are all positive odd integers, and $k\ge 5$, so $x_{i+1}$ is a positive odd integer, and $$y_{i+1}=(k-1)(k-3)y_i+2k-x_i\equiv -x_i\equiv 1(\mathrm{mod}2),$$ $y_{i+1}\ge 8y_i+2k-x_i>y_i>0$. Thus $(x_{i+1},y_{i+1})$ is one positive odd solution of ①, and $x_i+y_i<x_{i+1}+y_{i+1}$. By $(x_1,y_1)$ and the construction above, we get a series of positive odd solutions of ①: $(x_i,y_i)$, $i=1,2,\dots$, such that $x_1+y_1<x_2+y_2<\dots$. For any integer $i$ greater than $k$, $x_i+y_i>k$. Let $n=x_i+y_i-k$, $d_1=x_i$, $d_2=y_i$, then $n$ is a positive odd integer, and $d_1+d_2=n+k$. Since $(k-2{)}^2+1$ is even, we can show that $d_1,d_2$ are both divisors of $\frac{n^2+1}{2}$, and $d_1+d_2=n+k$. Thus such $n$ satisfies all the conditions, therefore there exist infinitely many positive odd integers $n$ satisfying the conditions.