Show that there are no 2-tuples $ (x,y)$ of positive integers satisfying the equation $ (x+1) (x+2)\cdots (x+2014)= (y+1) (y+2)\cdots (y+4028).$
Problem
Source: China Nanjing , 23 Mar 2014
Tags: number theory proposed, number theory, Diophantine equation, China TST
07.04.2014 18:40
ideas Remark that $x>2^{1200}-2014$ as the number of factors $2$ in the $RHS$ is at least $2014+1007+503$ . If $x=y^2+4029y+2705137$ the RHS is smaller, for $x>y^2+4029y+2705138$ the RHS > LHS.
01.02.2015 18:12
Does anyone have a solution?
08.03.2015 17:35
Could a generalization to this problem be: determine all positive integers $n$ such that $ (x+1) (x+2)\cdots (x+n)= (y+1) (y+2)\cdots (y+n)$ has a solution? However, for the China TST problem, would it help to look at the exponent of specific primes $p$ (e.g. 2,3,5, 4027) in the LHS and the RHS and forcing them to equate by setting the $x+i$'s to be divisible by $p^{(...)}$?
19.03.2015 01:15
Just wondering, does the China Math Olympiads have an official website? Thank you very much!
19.03.2015 17:27
probably not. most problems I find are on blogs or online forums.
12.04.2016 06:43
Did anyone get this problem during the test? I am very curious to see a solution to this beautiful and difficult problem!
12.04.2016 06:45
You do realize that very few people in the world could do this under timed conditions.
15.04.2016 04:55
Does anyone have a solution to this problem? Thank you very much to all your help!
15.04.2016 05:53
MathPanda1 wrote: Does anyone have a solution to this problem? Thank you very much to all your help! I have a solution to this problem! But I don't really want to help
15.04.2016 05:57
rmrf wrote: MathPanda1 wrote: Does anyone have a solution to this problem? Thank you very much to all your help! I have a solution to this problem! But I don't really want to help why?
15.04.2016 06:03
Hello, he lives in USA, has USAMO in 5 days, does not want spend energy typing solution
15.04.2016 07:47
qwerty137 wrote: Hello, he lives in USA, has USAMO in 5 days, does not want spend energy typing solution rmrf wrote: MathPanda1 wrote: Does anyone have a solution to this problem? Thank you very much to all your help! I have a solution to this problem! But I don't really want to help What was the point of writing this then?
15.04.2016 13:22
That's a signal for you solution will be in approximately 1 week
28.04.2016 02:59
Now that USA(J)MO is finished, could you post your solution please, rmrf? Thank you very much for all your help!
24.05.2016 22:06
Anyone? Thanks!
25.05.2016 07:42
Dear MathPanda1, It seem you ask many question but do not answer when other asked your. Always looking for solution to problems! Maybe slow down and make sure you can solving the problems you get solution to. You can start with this problem here
01.09.2016 04:28
Alright MathPanda1, if you're still interested in a complete solution (after about 3 months), here you go.
01.09.2016 04:43
Thank you so much for posting a complete solution to this difficult problem qwerty137! I really appreciate it.
19.02.2017 06:05
@MathPanda1, here is another solution, as promised.
09.10.2017 05:17
08.06.2018 20:20
MathPanda1 wrote: Could a generalization to this problem be: determine all positive integers $n$ such that $ (x+1) (x+2)\cdots (x+n)= (y+1) (y+2)\cdots (y+n)$ has a solution? However, for the China TST problem, would it help to look at the exponent of specific primes $p$ (e.g. 2,3,5, 4027) in the LHS and the RHS and forcing them to equate by setting the $x+i$'s to be divisible by $p^{(...)}$? Why you write $n$ to both side ? If you really want to look for an generalization replace $n$ by $2n$ at one side of the equality !
09.06.2018 04:04
neel02 wrote: MathPanda1 wrote: Could a generalization to this problem be: determine all positive integers $n$ such that $ (x+1) (x+2)\cdots (x+n)= (y+1) (y+2)\cdots (y+n)$ has a solution? However, for the China TST problem, would it help to look at the exponent of specific primes $p$ (e.g. 2,3,5, 4027) in the LHS and the RHS and forcing them to equate by setting the $x+i$'s to be divisible by $p^{(...)}$? Why you write $n$ to both side ? If you really want to look for an generalization replace $n$ by $2n$ at one side of the equality ! Unnecessary comment (like this one), obviously it's a typo.
25.10.2020 10:51
Solved with nukelauncher. First we show \(x\) is large. We analyze the 2-adic valuations of both sides. Observe from \(\left\lceil\frac{2014}{2^k}\right\rceil\le\frac{2014}{2^k}+1\) for \(k\le\log_2(x+2014)\) that the left-hand side has at least \[\sum_{k=1}^{\log_2(x+2014)}\left\lceil\frac{2014}{2^k}\right\rceil\le2014+\log_2(x+2014).\]factors of 2. Analogously, from \(\left\lfloor\frac{4028}{2^k}\right\rfloor\ge\frac{4028}{2^k}-1\) for \(k\le\log_24028\), the right-hand side has at most \[\sum_{k\ge1}\left\lfloor\frac{4028}{2^k}\right\rfloor\ge4028-\log_24028\]factors of 2. Combining these bounds, \[\log_2(x+2014)\ge2014-\log_24028\implies x\ge2^{2002}-2014,\]so \(x\) is sufficiently large. \bigskip Now we will bound \(x\) between two integers; in particular, we will show \(y^2+4029y+2705137<x<y^2+4029y+2705138\). Consider the given equation in the form \[(x+1)\cdots(x+2014)=\left(y^2+4029y+1\cdot4028\right)\cdots\left(y^2+4029y+2014\cdot2015\right).\]It is clear \(y^2+4029y+4027\le x\le y^2+4029y+2014^2\) by comparing smallest/largest terms. Set \(z=y^2+4029y\) and \(K=x-z\), so that \(4027\le K\le2014^2\); then \(z\ge2^{2000}\) is also very large. Consider the \(z^{2014-t}\) coefficient of \[P(z):=(z+K+1)\cdots(z+K+2014)-(z+1\cdot4028)\cdots(z+2014\cdot2015).\]Its magnitude is upper bounded by \[\binom{2014}t\Big[(K+2014)^t+(2014\cdot2015)\Big]\ll2^{200t}<z^{t/10}.\]Hence the \(z^{2014-t}\) monomial has magnitude at most \[z^{2014-t}\cdot z^{t/10}\ll z^{2013}/2013\]for \(t\ge2\). It follows that \(P(z)\) is dominated by the \(z^{2013}\) monomial. But for \(P\) to have a root \(z\ge2^{2000}\), the \(z^{2013}\) monomial must vanish to zero. Comparing the \(z^{2013}\) coefficients, we have \begin{align*} (K+1)\cdots+(K+2014)&=1+4028+\cdots+2014\cdot2015\\ \implies K&=2705137.5, \end{align*}which is not an integer, contradiction.
08.01.2022 07:30
Solved with rama1728. There must be at least $$\left \lfloor \frac{4028}{2^1} \right \rfloor+\left \lfloor \frac{4028}{2^2} \right \rfloor+\left \lfloor \frac{4028}{2^3} \right \rfloor +\dots = 4019$$powers of $2$ dividing the RHS. Besides the highest $v_2$ product term on the LHS, there can be at most $$\left \lceil \frac{2014}{2^2} -1 \right \rceil +\left \lceil \frac{2014}{2^2} -1 \right \rceil +\left \lceil \frac{2014}{2^3} -1 \right \rceil +\dots = 2004$$powers of $2$ dividing the LHS. So $x \ge 2^{2015}-2014.$ Let $A = y^2+4029y,$ then $$(x+1)\cdots (x+2014)= (A+1 \times 4028)\cdots (A+2014 \times 2015).$$Let $x - A = D.$ Note $A + 1 \times 4028 < x+1 \implies 4027 < D,$ and $x+2014 < A+2014\times 2015 \implies D < 2014^2.$ Some silly bounding follows. Consider the polynomial $P(a)$ with degree at most $2013$ and with integer coefficients where $$P(a) = (a+D+1) \cdots (a+D+2014) - (a+1 \times 4028) \cdots (a+2014 \times 2015),$$and $P(a)$ has a root $A.$ It suffices to prove that the $a^{2013}$ term is zero, since that would mean $D$ is not an integer for parity reasons. Consider the $a^{2014-n}$ term for any $2 \le n \le 2014,$ there are only $2013$ such terms in the polynomial. If we plug in $A,$ the term has magnitude (by Triangle Inequality) less than \begin{align*}\binom{2014}n \times \Big( (D+2014)^n+(2014\times 2015)^n\Big) \times A^{2014-n} &< 2048^n \times (2 \times 2048^{2n}) \times A^{2014-n}\\ &= \frac{2^{33n+1}}{A^{n-1}} \times A^{2013} \end{align*}which is much less than $\frac{A^{2013}}{2013}$ so the $a^{2013}$ coefficient has to be $0$ if $A$ is a root. $\square$
13.01.2024 17:00
It seems that no one has yet noted the similarity of this problem and the problem from the 239 MO in 2009 (author is Fedor Petrov). If I'm mistaken, please, say it...
16.01.2024 11:17
Ah, I remember this problem about two boys multiplying consecutive numbers, but I did not know their names until today.