An integrality question about expressing an integer as a product of numbers below $n$

Question

Let $n\ge 2$ be a natural number. Suppose that $N$ is a natural number, composed only of primes below $n$, and that can be expressed as $$ N= \prod_{j=1}^{n} j^{x_j} $$ where $x_1$, $\ldots$, $x_n$ are non-negative rational numbers with $\sum_{j}x_j \in {\Bbb N}$. Does there necessarily exist a representation of $N$ as $$ N=\prod_{j=1}^{n} j^{a_j} $$ where $a_1$, $\ldots$, $a_n$ are non-negative integers with $\sum_{j} a_j=\sum_j x_j$?

Let me also give the interpretation in terms of lattice points in polytopes in case that is easier to think about. For each natural number below $n$ associate a point in ${\Bbb Z}^{\pi(n)}$ corresponding to the exponents in the prime factorization of $n$. So $1$ corresponds to $(0,0,\ldots,0)$, $2$ corresponds to $(1,0,\ldots,0)$ and so on. Let ${\mathcal C}(n)$ denote the convex hull of these lattice points. Suppose we know that a number $N$ corresponds to a lattice point in ${\mathcal C}(n)$ dilated by a factor $m \in {\Bbb N}$. Can that lattice point then be written as a sum of $m$ lattice points from ${\mathcal C}(n)$? In general it is not true that the lattice points in the dilate of a set must fall in the $m$-fold sumset. Is it true in this special situation?

Note: I had originally overlooked the condition that $N$ should be composed only of primes below $n$. This subtlety was pointed out by The Masked Avenger, and at the same time Greg Martin produced an excellent example to show why it is necessary. Thanks to both of them, and with apologies to Greg for the oversight. Greg also provided a solution even when $N$ is composed only of primes below $N$. Here the problem is that the situation for numbers permits more relations than the situation for polytopes which is what I kept thinking about. So I've moved the goalpost one more time (sorry) by asking whether rationality of $x_j$ implies integrality. If this also has a simple counterexample, then I give up!

(Note: the condition that $\sum_j x_j \in {\Bbb N}$ is necessary, else as Christian Elsholtz kindly pointed out one has the counterexample $1^0\times 2^{1/2}\times 3^0\times 4^{1/4}=2$.)

This problem arose in connection with the recent MO question: How many different numbers can be obtained as product of first $n$ natural numbers? If the problem posed here has a positive answer, then it would follow (in the notation of the linked question) that $P(m,n)$ is the Erhart polynomial (in $m$) of a certain convex polytope ${\mathcal C}(n)$.

Answer 1

It seems that here is an example for rational exponents.

Let $n=2209=47^2>3^7>2^{11}$, and $N=4\,385\,664=2^7\cdot 3^6\cdot 47=2048^{7/11}\cdot 2187^{6/7}\cdot 2209^{1/2}\cdot 1^{1/154}$ with $7/11+6/7+1/2+1/154=2$. If $N=ab$ with $a,b\leq 47^2$, then 47 divides one of $a$ and $b$ (say, $a=47k$); since $a,b\leq 47^2$ we have $47\geq k\geq 40$ (the right estimate is quite rough). But there is no product of powers of 2 and 3 in this interval.

Answer 2

No. Set $$ x=\frac32, \quad y=\frac{\log(32/25)}{\log(16/9)}, \quad\text{and }z=\frac{\log(25/24)}{\log(16/9)}. $$ Then $2^x 3^y 4^z = 5$ and $x+y+z=2$, yet there is no way to write $5$ as a product of integers not exceeding $4$.

This might seem like a crazy coincidence, but it's not: for any integers $a<b<c<d$, the set of $(x,y,z)\in\mathbb R^3$ such that $$ a^x b^y c^z = d \quad\text{and}\quad x+y+z=2 $$ is the solution of a system of two linear equations. (Really! Take logarithms of the first one....) And under mild conditions (perhaps simply $c>\sqrt d$), the set of solutions will contain $(x,y,z)$ that are all nonnegative.

The same method can produce counterexamples to the updated claim as well: note that $48$ is composed only of primes not exceeding $7$, but cannot be written as the product of $2$ integers not exceeding $7$. Yet $2^u4^v7^w = 48$ where \begin{align*} u&=\frac{-110 \log 2-28 \log 3+55 \log 7}{28 \log 2}\\ v&=\frac{111 \log 2+28 \log 3-55 \log 7}{28 \log 2}\\ w&=\frac{55}{28} \end{align*} sum to $2$ and are all positive.