Is there an easy proof of the following statement?

$\forall$ $n>0 \in \mathbb N$, $ \exists$ $a\geq0 \in \mathbb N$ such that for any set of integers $(x_1,...,x_n)$ and $1\leq x_i \leq n$:

$(x_1,\dotsc,x_n)$ 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 $a(n)$?

**Extension:**

lambda's proof, using the Chinese remainder theorem allows a more general statement:

Notation: for an integer $m\geq1$ let's denote the set of integers $I_m=(1,2,...,m)$

The extension of the property:

$\forall$ $n>0 \in \mathbb N$, $ \exists$ $a\geq0 \in \mathbb N$ such that:

$\forall$ $p \in I_n$ ,

and for any sequences of integers $(x_1,...,x_p)$ and $(y_1,...,y_p)$ ; $x_i\in I_n$; $y_i\in I_n$,

If: $$\prod_{i=1}^p (a + x_i) = \prod_{i=1}^p (a + y_i)$$

then $(x_1,...,x_p)$ is a permutation of $(y_1,...,y_p)$

i.e. there exists a permutation $\sigma\in S_p $ such that : $x_i=y_{\sigma({i})}$ for $i=1,...,p$

I calculated the minimal $a$ for $n=1,2,3,4,5,6,7,8,9,10,11,12,13$ and got the values $a=0,0,0,1,2,5,6,9,10,16,27,27,28$

We can note that this new sequence of minimal $a$ differs from the previous one at least for $n=11$

($a=27$ instead of $18$)

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