, 2007

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 $x > y \geq 1$, $x!$ is divisible by $y!$, and $x! - y!$ is divisible by $y!$.

Let us consider the case $y \geq 2007$.
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 $x > y \geq 2007$, $y! > x^{2007}$ (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: $1 \leq y < 2007$.
For each such $y$, $x > y$ is a positive integer.
Let us check for small values of $y$:

If $y = 1$:
$$x^{2007} + 1! = 1^{2007} + x! \implies x^{2007} + 1 = 1 + x! \implies x^{2007} = x!.$$
But for $x = 2$, $2^{2007}$ is much larger than $2! = 2$.
For $x \geq 3$, $x!$ 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 $y = 2$:
$$x^{2007} + 2! = 2^{2007} + x! \implies x^{2007} + 2 = 2^{2007} + x! \implies x^{2007} - x! = 2^{2007} - 2.$$
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 $y = 3, 4, \ldots, 2006$, $x^{2007} - x!$ is always much larger than $y^{2007} - y!$ for $x > y$.

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

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