65th Czech and Slovak Mathematical Olympiad

Let us denote $v_p(n)$ the highest power of a prime $p$ which divides positive integer $n$. This function has obviously the following properties: - For all primes $p$ and positive integers $n$ is $v_p(n)$ non-negative integer. - For all positive integers $m,n$ and all primes $p$ is $v_p(mn)=v_p(m)+v_p(n)$. - For all primes $p$ is $v_p(p!)=v_p(p)=1$. - For all primes $p$ is $v_p((p+1)!)=1$, $v_p(p+1)=0$. - For all primes $p$ and all positive integers $n<p$ is $v_p(n!)=v_p(n)=0$. - Positive integer $n$ is a square if and only if $v_p(n)$ is even (including zero) for all primes $p$.

Let us denote $S=n!$ the initial value of the product on the table and $S^{\prime }$ its final value after adding factorials. We can easy to see from the properties of $v_p$ that for $n$ is equal to any prime $p$ we obtain $v_p(S)=v_p(p!)=1$ and $v_p(S^{\prime })=1$, because adding factorials does not change the amount of the prime $p$ (= $n$) in the final product on the blackboard. The number $v_p(S^{\prime })$ is then odd and therefore $S^{\prime }$ is not a square.

Let us assume that $n$ is a composite number (so $n\ge 4$) in whole of the following part. We will show that we can add factorials in such a way that the final product $$S^{\prime }=f_1\cdot f_2\cdot f_3\cdots f_n,$$ will be a square, where $f_k$ is either $k$ or $k!$ for all $k$. It is equivalent to $v_p(S^{\prime })$ is even for all primes $p$. Since $n$ is not a prime, only primes less than $n$ occur in the product $S^{\prime }$. As every such primes $p$ are not in factors $f_1,f_2,\dots ,f_{p-1}$ and the prime $p$ occurs in $f_p$ only once, the final power $v_p(S^{\prime })$ is the same as in a “reduced” product $$p\cdot f_{p+1}\cdot f_{p+2}\cdots f_n.(1)$$ How can we provide that every prime $p<n$ will occur in the corresponding product (1) with even power? Since in the second factor $f_{p+1}$ from (1) occurs the prime $p$ either once (in the case $f_{p+1}=(p+1)!$) or the prime $p$ does not occure (if $f_{p+1}=p+1$), we can provide “good” occurrence of $p$ by choice of $f_{p+1}$ independently on succeeding values $f_{p+2},\dots ,f_n$.

Foregoing analysis gives us construction of the required choice of factorials. Initially we choose $f_k\in \{k,k!\}$ arbitrarily for all $k\le n$ such that $k-1$ is not a prime. The other $f_k$, it is $f_{p+1}$, where $p$ is arbitrary prime less than $n$, will be chosen “backwards”, from the biggest such prime $p$ to the smallest prime $p=2$. For the biggest unchosen $f_{p+1}$ we find parity of $v_p(f_{p+2}\dots f_n)$, in odd case we choose $f_{p+1}=p+1$, in even case we choose $f_{p+1}=(p+1)!$ and so on.

This finishes the construction of $S^{\prime }$ and solution of the problem too.

Conclusion. Desired $n\ge 2$ are all composite numbers.