I have a question about an equality involving products of central binomial coefficients. If $x\_1,...,x\_n$ and $y\_1,...,y\_n$ are positive integers, with $\sum\_i x\_i = \sum\_i y\_i$ and 
$$ \binom{2x\_1}{x\_1} \cdots \binom{2x\_n}{x\_n} = \binom{2y\_1}{y\_1}\cdots \binom{2y\_n}{y\_n}\,, $$
what are the restrictions on the $x\_i$ and $y\_i$, and is there any solution other than the trivial one $\{x\_1,...,x\_n\}=\{y\_1,...,y\_n\}$?



Product of central binomial coefficients
Yes, there are nontrivial solutions. The first I found, with $n=5$, has
$$
\lbrace x\_i \rbrace = \lbrace 2, 5, 8, 13, 19 \rbrace,
\phantom{\infty}
\lbrace y\_i \rbrace = \lbrace 3, 4, 6, 14, 20 \rbrace,
$$
with $\sum\_i x\_i = \sum\_i y\_i = 47$ and
$$
\prod\_{i=1}^5 {2x\_i \choose x\_i} = \prod\_{i=1}^5 {2y\_i \choose y\_i}
 = 7153522697506948963200000
 = 2^{10} 3^6 5^5 7^3 11^2 13^1 17^1 19^1 23^2 29^1 31^1 37^1.
$$
This is a "list anagram" problem in multiplicative disguise; 
for another such example see puzzle #12
[here](http://math.harvard.edu/~elkies/Misc/index.html#puzzles)
and the notes and links in the
[solution](http://math.harvard.edu/~elkies/Misc/sol12.html) page.


[**EDIT**] I see that the OP did not specify that the $x\_i$ and $y\_i$
be distinct, which makes it a simpler problem (only linear algebra
over the rationals, no lattice reduction). But the version with
distinct variables is more appealing.




---


*[added later]* to spell it out: let $C(m) := {2m \choose m}$,
and suppose all $x\_i,y\_i$ are at most $M$.
For $1 \leq m \leq M$ let $a\_m = \#\{i: x\_i=m\} - \#\{i: y\_i=m\}$.
Then $x\_i,y\_i$ are a solution **iff**
$\sum\_{m=1}^M a\_m = \sum\_{m=1}^M m a\_m = 0$ and $\prod\_{m=1}^M C(m)^{a\_m}=1$.
By unique factorization, the last condition is equivalent to
$\sum\_{m=1}^M v\_p(C(m)) a\_m = 0$ for all primes $p$
(where $v\_p$ is the $p$-valuation), and we need only consider $p \leq 2M$.
So we have $\pi(2M)+2$ linear equations in $M$ variables.
There are nontrivial solutions provided $\pi(2M)+2 < M$,
which happens once $M$ is large enough because $\pi(x) = o(x)$;
it turns out that any $M>10$ is large enough.
The equations have integer coefficients, so any solution is rational,
and can be made integral by multiplying by a common denominator.
For example, the 1-dimensional space of solutions for $M=11$
is generated by $(a\_1,a\_2,\ldots,a\_{11}) = (1,-2,0,0,2,-1,0,0,0,1,-1)$,
giving $n=4$, $x\_i=1,5,5,10$, and $y\_i=2,2,6,11$ with a common sum of $21$
and
$$
\prod\_i C(x\_i) = \prod\_i C(y\_i) = 23465490048 = 2^7 3^4 7^2 11^1 13^1 17^1 19^1.
$$
The $x\_i$ and $y\_i$ are distinct **iff** the $a\_m$ are all $0$ or $\pm 1$.
There's no easy criterion for this, but it's correlated with small norm
$\sum\_{m=1}^M |a\_m|^2$, so we'll often find an example by applying LLL
(or some other technique for lattice basis reduction)
to the lattice of integer solutions of our linear system equation.
Fortunately the **matkerint** function of **gp** does this automatically,
and then **vecmax(abs(...))** can be used to detect vectors with all
coordinates $0$ or $\pm 1$. A bit more work finds such a vector
already for $M=18$, with $n=6$ and
$x\_i,y\_i = \lbrace 3, 6, 8, 11, 13, 17\rbrace,
\lbrace 4, 5, 7, 10, 14, 18 \rbrace$.


