Mfcttrf

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.

Here is a string of comments which might be helpful.

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  Consider instead cases of $$prod_1^k(x_i+a)= prod_1^k(y_i+a) tag*$$ where the multisets $x_1,cdots ,x_k$ and $y_1,cdots ,y_k$ are disjoint. I'll assume the elements are listed in increasing order. To stick to the OP, add the requirement that the $y_i$ are distinct. For example, $a(5)geq 2$ because there are counter-examples to $a=0$ and $a=1.$
  $$(2+0)(2+0)(3+0)(2+0)(5+0)=(1+0)(2+0)(3+0)(4+0)(5+0)$$
  $$(2+1)(2+1)(3+1)(3+1)(4+1)=(1+1)(2+1)(3+1)(4+1)(5+1)$$ Cancel out common factors to to see that sources of these counter-examples are $1cdot 4=2 cdot 2 $ and $2 cdot 6=3 cdot 4.$ In the other direction, one can pad an example of $(*)$ by changing the right-hand side to $prod_1^n(i+a)$ and adding on the left the same new factors. Here $n$ could be $y_k$ or anything larger.

  
  
  

  • The final remark exhibits that $a(n)$ is non-decreasing. So far $a(n) leq n+1$ although there seems no reason to conjecture that that continues.

  
  

  • Here is a potential conjecture. It is false. I mention it only because the counter-examples are lovely.

  
  

Suppose that the value of $prod_i=1^n (a + x_i) -prod_i=1^n (a + y_i)$ is independent of $a$. Does that mean that the shared value is $0$ and $x_i=y_i?$

The answer is no because of ideal solutions to the Prouhet-Tarry-Escott problem. For example $2^k+3^k+7^k=1^k+5^k+6^k$ for $k=0,1,2.$ This explains the observation that $$(2+a)(3+a)(7+a)=42+41a+12a^2+a^3$$
$$(1+a)(5+a)(6+a)=30+41a+12a^2+a^3$$