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Given two finite ordered sequences with possibly non-unique elements all greater than one: $A,B \in \mathcal{Z}_{>1}$. Given that we have:

\begin{eqnarray} |A| &= |B| \\ \Pi_{x \in A} x &= \Pi_{x \in B} x \\ \Sigma_{x \in A} x &= \Sigma_{x \in B} x \end{eqnarray}

Is it true that $B$ must be a permutation of $A$?

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up vote 4 down vote accepted

No. Take $A = \{ 2, 10, 9, 12 \}$ and $B = \{ 4, 5, 6, 18 \}$.

Here is how I constructed the above example. I found a number, namely $12$, which admits two factorizations $$12 = 2 \cdot 6 = 3 \cdot 4$$

such that the sum of the numbers used in each factorization differs by a small amount (in this case $1$). I found a second such number, namely $20$: $$20 = 4 \cdot 5 = 2 \cdot 10.$$

In this case the sum of the numbers used in each factorization differs by $3$. If I multiply each of the numbers in the first example by $3$, I get a second number $$108 = 6 \cdot 18 = 9 \cdot 12$$

with two factorizations such that the sum of the numbers used in each factorization differs by $3$. By mixing these I can arrange for the sums to be equal, and by construction the products are clearly equal.

Generalizing the above argument yields the $6$-parameter family of examples $$A = \{ ta, tbc, sde, sf \}, B = \{ tab, tc, sd, sef \}$$

where $t = (de + f) - (d + ef)$ and $s = (ab + c) - (a + bc)$. (I may have a sign wrong.) One obtains the above example by setting $a = 3, b = c = 2, d = 5, e = f = 2$.

Here is a counterexample with triplets: $A = \{ 2, 8, 9 \}, B = \{ 3, 4, 12 \}$. To get these I started with numbers of the form $$A = \{ a, bc, de \}, B = \{ ab, cd, e \}$$

which have the same product; demanding they have the same sum gives $$a + bc + de = ab + cd + e.$$

Setting $b = 2$ gives $$a = e(d-1) - c(d-2)$$

and yields a $3$-parameter family of solutions in the parameters $c, d, e$. The above example is obtained by setting $d = 3, c = 4, e = 3$.

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Thanks Qiaochu, the construction was far more useful than a simple counter-example. – Hooked May 22 '12 at 15:55
Yes, but the construction took longer to explain! In fact there was an error in the counterexample which has been fixed. Note that in general you can let $b$ and $e$ be rational and still get integer examples in the above. – Qiaochu Yuan May 22 '12 at 16:09

There is a symmetric system of algebraic conditions that do work. Suppose that $|A|=|B|=n$.

The $k$th elementary symmetric polynomial $e_k(x_1,\cdots,x_n)$ is defined as

$$e_k:= \sum_{1\le j_1\le\cdots\le j_k\le n}x_{\large j_1}x_{\large j_2}\cdots x_{j_k}.$$

In particular, $e_1$ is the sum of the terms while $e_n$ is the product, as in your two conditions. Then two sequences of numbers $(a_1,\cdots,a_n)$ and $(b_1,\cdots,b_n)$ are permutations of the other if and only if

$$e_k(a_1,\cdots,a_n)=e_k(b_1,\cdots,b_n)\quad \text{for each }k=1,2,\cdots,n.$$

This is because $p(x)=(x-a_1)\cdots(x-a_n)$ and $q(x)=(x-b_1)\cdots(x-b_n)$ are the same integer-coefficient polynomials if and only if the roots $\{a_i\}$ and $\{b_i\}$ are the same (even up to multiplicity, so the terms in the sequence don't necessarily need to be distinct). At the same time, by Vieta's formulas, we have the power expansions

$$p(x)=x^n-e_1(a_1,\cdots,a_n)x^{n-1}+\cdots\mp e_{n-1}(a_1,\cdots,a_n)x\pm e_n(a_1,\cdots,a_n); $$

$$q(x)=x^n-e_1(b_1,\cdots,b_n)x^{n-1}+\cdots\mp e_{n-1}(b_1,\cdots,b_n)x\pm e_n(b_1,\cdots,b_n).$$

Two polynomials are equal if and only if all of their coefficients are equal, so comparing coefficients gives the result.

If we work with sequences of complex numbers, it is even possible to a priori prescribe the values of the $e_k$ and know that there will be a a set of roots to the corresponding polynomial, so it is possible to prescribe precisely what $e_k$ values two sets of numbers take - including where they differ - and it is gauranteed there will be two sets of numbers to fit the bill. Put these two sets into sequence form arbitrarily and we end up with two sequences of numbers with exactly the $e_k$ values we want, including when they are equal and when they aren't, that are not the same even up to permutation.

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