59th Ukrainian National Mathematical Olympiad

We will start with the following lemma. Lemma 1. There exists a diagonal that has two equal numbers for any $2\times 2$ square. Proof. Let $X$ be the greatest number in $2\times 2$ square. Let his neighbors in this $2\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if $A=B$, the lemma is proven. Suppose $A>B$. Then $A+X=k!>B+X=m!>1$, thus $k>m$. It is also clear that $m>1,k>2$. But then $$2(B+X)=2m!>2X\ge A+X=k!\ge k\cdot m!>2m!,$$ that leads to a contradiction. This finishes the proof of Lemma 1.

By contradiction, suppose $a,b,c,d,e,f,g,h,i$ are in the table (Fig. 5). By Lemma 1 for the square that consists of $a,b,d,e$ either $b=d$ or $a=e$. In the first case, we will use Lemma 1 for the square with $d,e,g,h$, and then for $b,c,e,f$. Thus, either $b=d=h$, or $b=d=f$, or $g=e=c$.

In the second case, we will use Lemma 1 for $e,f,h,i$, and then for $b,c,e,f$. Thus, either $a=e=i$ or $a=e=c$, or $h=f=b$. In all cases we obtain three equal numbers.