# Necessary and sufficient condition for $(x_1,...,x_n)$ to be a permutation of $(1,...,n)$ - Part II

This thread is following the question posted here and seeks to find a generalization to the test function used in the previous question i.e. $$f_a(x_1, \dotsc,x_n)=(x_1+a)\dotsb(x_n+a)$$

At first, I propose this conjecture:
$$\forall$$ $$n>0 \in \mathbb N$$, $$\exists$$ $$p>0 \in \mathbb N$$ 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:
$$x_1^p+\dotsb+x_n^p=1^p+\dotsb+n^p$$

I checked this conjecture for $$n=1,\dotsc,17$$
The minimum values of $$p$$ are: $$(1,1,2,2,2,4,5,5,5,5,5,7,7,7,7,7,7)$$
Here the test function is $$f_p(x_1, \dotsc,x_n)=x_1^p+\dotsb +x_n^p$$

The question is then: assuming $$f_p(x_1, \dotsc,x_n)$$ is a symmetric function of n variables with a parameter $$p$$, what kind of properties must $$f$$ have in order to be a test function i.e. to be able to say if $$(x_1,...,x_n)$$ is a permutation of $$(1,...,n)$$?
New examples ?

• I wonder what sets $S$ of integer partitions of $n$ has the property that if one knows the values of $s_\lambda(x)$ for $\lambda \in S$, then one can deduce 'permutability' of the vector $x$. Here, $s_\lambda$ are the Schur functions. Jul 18, 2019 at 6:44

Your conjecture is true: If $$p$$ is big enough then $$(n-1)\cdot (n-1)^p and therefore $$\lfloor \frac{x_1^p+\cdots+x_n^p}{n^p}\rfloor$$ denotes the number of $$i$$ with $$x_i=n$$, hence from $$x_1^p+\cdots + x_n^p=1^p+\cdots +n^p$$ we see that this number is one as desired. Next, if also $$(n-2)\cdot(n-2)^p<(n-1)^p$$, then $$\lfloor \frac{x_1^p+\cdots+x_n^p-n^p}{(n-1)^p}\rfloor$$ denotes the number of $$i$$ with $$x_i=n-1$$, and so on. So all we need is $$p$$ such that $$k^{p+1}<(k+1)^p$$ for $$1\le k\le n-1$$. Divide by $$k^p$$ and take logarithms to see that it is sufficient to take $$p>\max_{1\le k (This gives $$p=50$$ instead of $$p=7$$ for $$n=17$$, but who cares)
• Just one small question: if all $x_i$ are equal to $n-1$, we have $x_1^p + \dots + x_n^p= n \cdot (n-1)^p$. Hence I think the inequality $(n-1)\cdot (n-1)^p < n^p$ is formally not sufficient for the first step, I think for that we might need $n \cdot (n-1)^p < n^p$. Mar 20, 2021 at 18:50