China Mathematical Olympiad

Proof. Let $a_1,a_2,\dots ,a_{k+1}$ be $k+1$ distinct positive integers, where each of them is smaller than the product of other $k$ numbers. Write $N=a_1{a}_2\cdots a_{k+1}$. For each $i=1,2,\dots ,k+1$, let $x_i=\frac{1}{2}\left(\frac{N}{a_i}+a_i\right)$, $y_i=\frac{1}{2}\left(\frac{N}{a_i}-a_i\right)$, then $x_i^2-y_i^2=N$. Since $a_i{a}_j<N$ and $\frac{N}{a_i}>a_i$, $(x_i,y_i)(1\le i\le k+1)$ are $k+1$ positive integer solutions of equation $x^2-y^2=N$. Without loss of generality, suppose $x_{k+1}=\min \{x_1,x_2,\dots ,x_{k+1}\}$. For each $i\in \{1,2,\dots ,k\}$, since $x_i^2-y_i^2=x_{k+1}^2-y_{k+1}^2$, we have $$(x_i+x_{k+1})(x_i-x_{k+1})=x_i^2-x_{k+1}^2=y_i^2-y_{k+1}^2=(y_i+y_{k+1})(y_i-y_{k+1}).$$ Let $n_i=(x_i+x_{k+1})(x_i-x_{k+1})=(y_i+y_{k+1})(y_i-y_{k+1})$, then $$2x_{k+1}=(x_i+x_{k+1})-(x_i-x_{k+1})\in D(n_i),$$ $$2y_{k+1}=(y_i+y_{k+1})-(y_i-y_{k+1})\in D(n_i).$$ So $x_{k+1}>y_{k+1}$, $2x_{k+1}$ and $2y_{k+1}$ are two different members of $D(n_1)\cap D(n_2)\cap \cdots \cap D(n_k)$. $\square$