Subfield: Combinatorics


Question: A finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers



$$

a+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0)

$$



Prove that, for every integer $n>100$, there are only finitely many integers $k \geq 0$, such that, starting from the list



$$

k+1, k+2, \ldots, k+n

$$



it is possible to obtain, after $n-1$ operations, the value $n$ !.

Solution:
We prove the problem statement even for all positive integer $n$.



There are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elementary operation from two such formulas on two disjoint sets of the $x_{i}$.



A straightforward induction on $n$ shows that the outcome of each such construction is a number of the form



$$

\frac{\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} a_{\alpha_{1}, \ldots, \alpha_{n}} x_{1}^{\alpha_{1}} \cdots x_{n}^{\alpha_{n}}}{\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} b_{\alpha_{1}, \ldots, \alpha_{n}} x_{1}^{\alpha_{1}} \cdots x_{n}^{\alpha_{n}}},
\tag{*}
$$



where the $a_{\alpha_{1}, \ldots, \alpha_{n}}$ and $b_{\alpha_{1}, \ldots, \alpha_{n}}$ are all in the set $\{0, \pm 1\}$, not all zero of course, $a_{0, \ldots, 0}=b_{1, \ldots, 1}=0$, and also $a_{\alpha_{1}, \ldots, \alpha_{n}} \cdot b_{\alpha_{1}, \ldots, \alpha_{n}}=0$ for every set of indices.



Since $\left|a_{\alpha_{1}, \ldots, \alpha_{n}}\right| \leq 1$, and $a_{0,0, \ldots, 0}=0$, the absolute value of the numerator does not exceed $\left(1+\left|x_{1}\right|\right) \cdots\left(1+\left|x_{n}\right|\right)-1$; in particular, if $c$ is an integer in the range $-n, \ldots,-1$, and $x_{k}=c+k$, $k=1, \ldots, n$, then the absolute value of the numerator is at most $(-c) !(n+c+1) !-1 \leq n !-1<n !$.



Consider now the integral polynomials,



$$

P=\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} a_{\alpha_{1}, \ldots, \alpha_{n}}(X+1)^{\alpha_{1}} \cdots(X+n)^{\alpha_{n}}

$$



and



$$

Q=\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} b_{\alpha_{1}, \ldots, \alpha_{n}}(X+1)^{\alpha_{1}} \cdots(X+n)^{\alpha_{n}}

$$



where the $a_{\alpha_{1}, \ldots, \alpha_{n}}$ and $b_{\alpha_{1}, \ldots, \alpha_{n}}$ are all in the set $\{0, \pm 1\}$, not all zero, $a_{\alpha_{1}, \ldots, \alpha_{n}} b_{\alpha_{1}, \ldots, \alpha_{n}}=0$ for every set of indices, and $a_{0, \ldots, 0}=b_{1, \ldots, 1}=0$. By the preceding, $|P(c)|<n$ ! for every integer $c$ in the range $-n, \ldots,-1$; and since $b_{1, \ldots, 1}=0$, the degree of $Q$ is less than $n$.



Since every non-zero polynomial has only finitely many roots, and the number of roots does not exceed the degree, to complete the proof it is sufficient to show that the polynomial $P-n ! Q$ does not vanish identically, provided that $Q$ does not (which is the case in the problem).



Suppose, if possible, that $P=n ! Q$, where $Q \neq 0$. Since $\operatorname{deg} Q<n$, it follows that $\operatorname{deg} P<n$ as well, and since $P \neq 0$, the number of roots of $P$ does not exceed $\operatorname{deg} P<n$, so $P(c) \neq 0$ for some integer $c$ in the range $-n, \ldots,-1$. By the preceding, $|P(c)|$ is consequently a positive integer less than $n$ !. On the other hand, $|P(c)|=n !|Q(c)|$ is an integral multiple of $n$ !. A contradiction.