First selection test for IMO 2007, Vietnam

Assume, for the sake of contradiction, that there exist distinct positive integers $x$ and $y$ such that $$x^{2007}+y!=y^{2007}+x!.$$ Without loss of generality, suppose $x>y$.

Then, $$x^{2007}-y^{2007}=x!-y!.$$ Let us estimate the size of both sides for large $x$ and $y$.

Note that for $x>y\ge 1$, $x!$ is divisible by $y!$, and $x!-y!$ is divisible by $y!$.

Let us consider the case $y\ge 2007$. Then $y!$ is divisible by all numbers up to $y$, in particular by $y^{2007}$, so $y!$ is divisible by $y^{2007}$. But $x!-y!$ is divisible by $y!$, so $x^{2007}-y^{2007}$ is divisible by $y!$. But $x^{2007}-y^{2007}<x^{2007}$, and for $x>y\ge 2007$, $y!>x^{2007}$ (since factorial grows faster than any fixed power). Therefore, $x^{2007}-y^{2007}<y!$, so the only way $y!$ divides $x^{2007}-y^{2007}$ is if $x^{2007}-y^{2007}=0$, i.e., $x=y$, which contradicts the assumption that $x$ and $y$ are distinct.

Therefore, $y<2007$.

Now, $y$ can only take finitely many values: $1\le y<2007$. For each such $y$, $x>y$ is a positive integer. Let us check for small values of $y$:

If $y=1$: $$x^{2007}+1!=1^{2007}+x!⟹x^{2007}+1=1+x!⟹x^{2007}=x!.$$ But for $x=2$, $2^{2007}$ is much larger than $2!=2$. For $x\ge 3$, $x!$ grows much faster than $x^{2007}$ only for very large $x$, but for small $x$, $x^{2007}$ is much larger than $x!$. So there is no solution for $y=1$.

If $y=2$: $$x^{2007}+2!=2^{2007}+x!⟹x^{2007}+2=2^{2007}+x!⟹x^{2007}-x!=2^{2007}-2.$$ But for $x=3$, $3^{2007}$ is much larger than $3!=6$. For $x=4$, $4^{2007}$ is much larger than $4!=24$. So there is no solution for $y=2$.

Similarly, for $y=3,4,\dots ,2006$, $x^{2007}-x!$ is always much larger than $y^{2007}-y!$ for $x>y$.

Therefore, there are no solutions in positive integers $x$ and $y$ with $x\ne y$.

Thus, there are no distinct positive integers $x$ and $y$ such that $$x^{2007}+y!=y^{2007}+x!.$$