show that this equation $$(x+1)(x+2)(x+3)\cdots(x+2014)=(y+1)(y+2)(y+3)\cdots(y+4028)$$ have no positive integer solution.
This problem is china TST (2014),I remember a famous result? maybe is a Erdos have proof this problem?
maybe this follow have some paper? $$(x+1)(x+2)(x+3)\cdots(x+k)=(y+1)(y+2)(y+3)\cdots(y+2k)$$ positive integer solution?
Thank you for help
On
This is not an answer, but it's too long for a comment. Perhaps it helps:
Observe that $$\frac{(y+1)\cdots(y+4028)}{4028!}={y+4028\choose y},$$ which is an integer. Then $$(x+1)\cdots(x+2014)=4028!{y+4028\choose y},$$ so the left hand side is an integer multiple of $4028!$.
This yields a large lower bound for $x$ as follows. The product $(x+1)\cdots(x+2014)$ must be divisible by the product of $p$, $2p$, and $3p$, for any prime with $3p\le4028$. If $p>1007$, there isn’t room for three multiples of $p$ between $x+1$ and $x+2014$, so one of the $x+j$ must be a multiple of $p^2$. Using $p=1327$, then $x>1327^2$.
You can also get some partial congruences for $x$. The prime $4027$ divides $(x+1)\cdots(x+2014)$, so $(x \mbox{ mod } 4027)\ge2013$.
[Note that in my comment to the original question, I provide a reference to a general result implying that there is no solution to the original equation.]
On
The $\nu_2$ (the exponent of the greatest power of $2$ dividing) of the RHS is at least $4010$, so there must be an element $\xi\in X=\{x+1,\ldots,x+2014\}$ such that $\nu_2(\xi)\geq 1996$. This gives that, if a solution $(x,y)$ exists, it must be huge, with $x>2^{1995}$ and $y$ around $\sqrt{x}$. For the same reason (with $\nu_3$ in place of $\nu_2$), if a solution exists, a multiple of $3^{995}$ must lie in $X$. So many multiples of huge powers of different primes must lie in a relatively short interval: this is quite uncommon, since it gives that $$ \Omega\left(\prod X\right) $$ is extremely big.
We also have that all the primes in the $[2015,4028]$ interval divide $\prod X$: $$2017,2027,2029,2039,\ldots,4027\; | \prod X\tag{2},$$ so, additionally, $x$ can lie in a few residue classes $\pmod{\prod_{2014<p<4028}p}$.
On
According to Theorem 4 in the paper "On the ratio of two blocks of consecutive integers," the equation $$(x+1)\cdots(x+k)=(y+1)\cdots(y+2k)$$ has only one solution in integers $x\geq0,y\geq0,k\geq2$ and it is given by $x=7,y=0,k=3$.
OK this doesn't quite work, but here's some ideas.
Let $k=2014$ for ease of notation. Suppose we have a solution.
(1) Expand the RHS as $\prod_{i=1}^{k} (y^2+(2k+1)y+i(2k+1-i))$. We see that $y^2+(2k+1)y+2k<x<y^2+(2k+1)y+k^2$.
(2) Suppose for a prime $p$ and $a\ge1$, $p^a$ is the largest power dividing any factor $y+i$ on the RHS. By (1), we have that each factor on the LHS is of the form $$y^2+(2k+1)y+c=(y+i)^2+(2k+1-2i)(y+i) +i^2-(2k+1)i+c$$ for $c\in(2k,k^2+2k)$. Letting $y'=y+i$, this is $$y'^2+(2k+1-2i)y'+d$$ for $d\in (-k^2,k^2)$. Now assuming $2k<k^2<p^a$, we have $d<p^a$ so the only way this is divisible by $p^a$ is if $d=0$; then the highest power of $p$ that can divide this is $p^{a+\log_{p}(2k)}$. Suppose the highest power of $p$ dividing a term on the LHS is $p^{a+j}$; we have $j\le \log_p (2k)$.
(3) Assuming $k^2<p^a$ we have $$ v_p(LHS)=\sum_{i=1}^{a+j}\left\lfloor{\frac{x+k}{p^i}}\right\rfloor -\left\lfloor{\frac{x}{p^i}}\right\rfloor $$ $$ v_p(RHS)=\sum_{i=1}^{a}\left\lfloor{\frac{y+2k}{p^i}}\right\rfloor -\left\lfloor{\frac{y}{p^i}}\right\rfloor $$ Using $x\ge \lfloor x\rfloor>x-1$ and noting that the terms in the sum for $\log_p(2k)<i\le a$ are all 1 in the 2nd sum and at most 1 in the first sum, $$ 0=v_p(RHS)-v_p(LHS)\ge \left[\sum_{i=1}^{\log_p (2k)}\left(\frac{2k}{p^i}-1\right) -\left(\frac{k}{p^i}+1\right)\right]-\log_p(2k)\ge \frac kp-3\log_p(2k). $$