Xjyuk

Is there an easy proof of the following statement?

$forall$ $n>0 in mathbb N$, $ exists$ $ageq0 in mathbb N$ such that
for any set of integers $(x_1,...,x_n)$ and $1leq x_i leq n$:

$(x_1,dotsc,x_n)$ is a permutation of $(1,dotsc,n)$ if and only if:
$(x_1+a)dotsb(x_n+a)=(1+a)dotsb(n+a)$.

I checked the property for $n=1,2,dotsc,9$ and got the (minimal) values $a=0,0,0,1,2,5,6,9,10$.

If the property is true, what can we say about the function $a(n)$?

Let $p_1, dots, p_n$ be distinct prime numbers each greater than $n$. By the Chinese remainder theorem, there exists an $a$ such that $a + i$ is divisible by $p_i$ for $1 le i le n$. Since $p_i > n$, it follows that if $1 le j le n$ and $p_i$ divides $a + j$ then $i = j$. In particular, if $(x_1, dots, x_n)$ lie in this range and $$prod_i=1^n (a + x_i) = prod_i=1^n (a + i)$$ then for each $i$ the product is divisible by $p_i$ so there is some $j$ such that $x_j = i$. Thus it's a permutation.

This proves existence but I'd expect the value of $a$ you get this way to be far from optimal.

Start by thinking about $prod_k=1^n (x_k+alpha)$ as the polynomial $f_x_1,dots,x_n(alpha)$ in $a$ with roots at $-x_1$,...,$-x_n$. Then the equality of polynomials
$$f_x_1,dots,x_n(alpha)=f_1,2,dots,n(alpha) (=:sum_k=0^n c_kalpha^k)quadtag1$$
holds iff $x_1,...,x_n$ is a permutation of $1,2,...,n$.
Now, in order to answer the 1st question it suffices to find a value $a$ of $alpha$ so that the equality of values of these polynomials at $a$ implies (1). That such $a$ exists follows from a standard argument involving thinking of a $sum_k=0^n b_k a^k$, with $a>max_k b_k$ as a number in base $a$.
Thus, it suffices to choose $a>max_k c_k$, with $c_k$ as in (1).

Finding out the minimal $a$ for each $n$ appears to be a much harder problem.