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, let the table consist of the numbers $a_1,a_2,...,a_{16}$, that satisfy the conditions (Fig. 7). Use Lemma 1 for a square with $a_1,a_2,a_5,a_6$. Without loss of generality, let $a_2=a_5$. Use Lemma 1 for squares with $a_2,a_3,a_6,a_7$ and $a_5,a_6,a_9,a_{10}$.

Case I. $a_2=a_7$ or $a_5=a_{10}$. Since these cases are similar, let $a_2=a_5=a_{10}=x$.

$a_1$	$x$	$a_3$	$a_4$
$x$	$a_6$	$a_7$	$a_8$
$a_9$	$x$	$a_{11}$	$a_{12}$
$a_{13}$	$a_{14}$	$a_{15}$	$a_{16}$

Fig. 8

From Lemma 1 for the square $a_9$, $a_{10}=x,a_{13},a_{14}$, if $a_{13}=x$, then we have four equal numbers that lead to a contradiction. Then $a_9=a_{14}$. Similarly, from Lemma 1 for a square with $a_{10}=x,a_{11},a_{14},a_{15}$ we have that $a_9=a_{14}=a_{11}=y$

$a_1$	$x$	$a_3$	$a_4$
$x$	$a_6$	$a_7$	$a_8$
$y$	$x$	$y$	$a_{12}$
$a_{13}$	$y$	$a_{15}$	$a_{16}$

Fig. 9

It suffices to use the Lemma 1 for a square with $a_6,a_7,a_{10}=x,a_{11}=y$ thus, the table has either four numbers $x$ or four numbers $y$. The contradiction completes the proof.

Case II. $a_3=a_6=a_9=t$.

$a_1$	$a_2$	$t$	$a_4$
$a_5$	$t$	$a_7$	$a_8$
$t$	$a_{10}$	$a_{11}$	$a_{12}$
$a_{13}$	$a_{14}$	$a_{15}$	$a_{16}$

Fig. 10

By assumption, there is no other number $t$ among the rest of the values, so by Lemma 1 the following holds: $a_4=a_7=a_{10}=a_{13}$, that leads to a contradiction and completes the proof.