Problem 1465

Given that $m$ and $n$ are positive integers with the property:
$$(mn) \mid (m^{2010} + n^{2010} + n)$$
we need to show that there exists a positive integer $k$ such that $n = k^{2010}$.

1. **Express $m$ and $n$ in terms of their greatest common divisor:**
Let $\gcd(m, n) = d$. Then we can write:
$$m = ad \quad \text{and} \quad n = bd$$
where $\gcd(a, b) = 1$.

2. **Substitute $m$ and $n$ into the given divisibility condition:**
$$(mn) \mid (m^{2010} + n^{2010} + n)$$
Substituting $m = ad$ and $n = bd$, we get:
$$(ad \cdot bd) \mid ((ad)^{2010} + (bd)^{2010} + bd)$$
Simplifying, we have:
$$(abd^2) \mid (a^{2010}d^{2010} + b^{2010}d^{2010} + bd)$$

3. Simplify the divisibility condition:
$$abd^2 \mid d^{2010}(a^{2010} + b^{2010}) + bd$$
Dividing both sides by $d^2$, we get:
$$ab \mid d^{2008}(a^{2010} + b^{2010}) + b$$

4. Analyze the divisibility:
Since $ab \mid d^{2008}(a^{2010} + b^{2010}) + b$, and $\gcd(a, b) = 1$, $a$ must divide $b$. Therefore, $a \mid b$.

5. **Consider the implications of $a \mid b$:**
Since $\gcd(a, b) = 1$, the only possibility is $a = 1$. Thus, $m = d$ and $n = bd$.

6. **Substitute $a = 1$ into the simplified condition:**
$$b \mid d^{2008}(1 + b^{2010}) + b$$
Simplifying further, we get:
$$b \mid d^{2008} + 1$$

7. **Conclude the form of $b$:**
Since $b$ must divide $d^{2008} + 1$, and $b$ is a positive integer, the simplest solution is $b = 1$. Therefore, $n = d^{2010}$.

8. **Define $k$:**
Let $k = d$. Then $n = k^{2010}$.

Thus, we have shown that there exists a positive integer $k$ such that $n = k^{2010}$.

$\blacksquare$