China Team Selection Test

Proof For $n=2^k\cdot m$ ($k$ is a non-negative integer, $m$ is odd), let $v(n)=2^k$. We prove by contradiction: assume $(x,y)$ is one positive integer solution of the equation. Let $$v(x+i)=\underset{1\le j\le 2014}{\max }\{v(x+j)\},$$ then if $1\le j\le 2014,j\ne i$, $$v(x+j)=v(x+i+(j-i))=v(j-i),$$ so $$v\left(\prod _{1\le j\le 2014,j\ne i}(x+j)\right)=v\left((2014-j)!\cdot (j-1)!\right)\le v(2013!)$$ since ${\prod }_{j=1}^{2014}(x+j)={\prod }_{j=1}^{4028}(y+j)$ is a multiple of $4028!$, thus $$x+i\ge v(x+i)\ge v\left(\frac{4028!}{2013!}\right)>2^{1007},$$ therefore $x>2^{1006}$. So $(y+4028{)}^{4028}>{\prod }_{j=1}^{4028}(y+j)={\prod }_{j=1}^{2014}(x+j)>2^{1006\cdot 2014}$, we have $y+4028>2^{503}$, $y>2^{502}$.

Lemma: Let $0\le x_i<\frac{1}{2}(1\le i\le n)$, if $x=\frac{1}{n}{\sum }_{i=1}^n{x}_i$, $y=2{\max }_{1\le i\le n}\{x_i^2\}$, then $$1-x\ge {\left(\prod _{i=1}^n(1-x_i)\right)}^{\frac{1}{n}}\ge 1-x-y.$$ Proof of lemma: By AM-GM inequality, the inequality on the left is easy to prove. The inequality on the right holds, since $$\begin{array}{rl}{\left(\prod _{i=1}^n(1-x_i)\right)}^{\frac{1}{n}}& \ge \frac{n}{{\sum }_{i=1}^n\frac{1}{1-x_i}}\\ & \ge \frac{n}{{\sum }_{i=1}^n(1+x_i+2x_i^2)}\\ & =\frac{n}{n+nx+2{\sum }_{i=1}^n{x}_i^2}\\ & \ge \frac{1}{1+x+y}\\ & \ge 1-x-y,\end{array}$$ the lemma is proved.

Back to the problem, Let $w=x+2015>2^{1007}$, $z=y+\frac{4029}{2}>2^{502}$, then the equation is equivalent to $$w\cdot {\left(\left(1-\frac{1}{w}\right)\left(1-\frac{2}{w}\right)\cdots \left(1-\frac{2014}{w}\right)\right)}^{\frac{1}{2014}}=z^2\cdot {\left(\left(1-\frac{1}{4z^2}\right)\left(1-\frac{9}{4z^2}\right)\cdots \left(1-\frac{4027^2}{4z^2}\right)\right)}^{\frac{1}{2014}}.$$ By lemma, $$w\left(1-\frac{2015}{2w}\right)>w\cdot {\left(\left(1-\frac{1}{w}\right)\left(1-\frac{2}{w}\right)\cdots \left(1-\frac{2014}{w}\right)\right)}^{\frac{1}{2014}}$$ $$\begin{array}{rl}& >w\left(1-\frac{2015}{2w}-\frac{2\cdot 2014^2}{w^2}\right)\\ & >w\left(1-\frac{2015}{2w}\right)-\frac{1}{8},\end{array}$$ therefore the decimal of $w \cdot \left( \left( 1 - \frac{1}{w} \right) \left( 1 - \frac{2}{w} \right) \cdots \left( 1 - \frac{2014}{w} \right) \right)^{\frac{1}{2014}}$ belongs to $\left( \frac{3}{8}, \frac{1}{2} \right)$. On the other hand, by lemma $$\begin{array}{rl}{z}^2\left(1-\frac{1^2+3^2+\cdots +4027^2}{4z^2\cdot 2014}\right)& >z^2\cdot {\left(\left(1-\frac{1}{4z^2}\right)\left(1-\frac{9}{4z^2}\right)\cdots \left(1-\frac{4027^2}{4z^2}\right)\right)}^{\frac{1}{2014}}\\ & >z^2\left(1-\frac{1^2+3^2+\cdots +4027^2}{4z^2\cdot 2014}-\frac{2\cdot 4027^4}{(4z^2{)}^2}\right),\end{array}$$ $$thus$$ $$\begin{array}{rl}{z}^2-\frac{4\cdot 2014^2-1}{12}& >z^2\cdot {\left(\left(1-\frac{1}{4z^2}\right)\left(1-\frac{9}{4z^2}\right)\cdots \left(1-\frac{4027^2}{4z^2}\right)\right)}^{\frac{1}{2014}}\\ & >z^2-\frac{4\cdot 2014^2-1}{12}-\frac{4027^4}{8z^2}\\ & >z^2-\frac{4\cdot 2014^2-1}{12}-\frac{1}{8}.\end{array}$$ Since $z^2 - \frac{4 \cdot 2014^2 - 1}{12}$ is an integer, so the decimal of $z^2 \cdot \left( \left( 1 - \frac{1}{4z^2} \right) \left( 1 - \frac{9}{4z^2} \right) \cdots \left( 1 - \frac{4027^2}{4z^2} \right) \right)^{\frac{1}{2014}}$ belongs to $\left( \frac{7}{8}, 1 \right)$, this is a contradiction. Therefore, there are no 2-tuples $(x, y)$ of positive integers satisfying \prod_{j=1}^{2014} (x + j) = \prod_{j=1}^{4028} (y + j). $$