Problem 1475

To prove that there exist pairwise distinct positive integers $a, b, c$ such that $n = ab + bc + ca$, we will use the given conditions and construct such integers.

1. **Assume $k < p$**:
Since $p$ is a prime and $k$ is an integer such that $p \mid (n + k^2)$, we can assume without loss of generality that $k < p$. If $k \geq p$, we can replace $k$ with $k \mod p$ because $p \mid (n + k^2)$ implies $p \mid (n + (k \mod p)^2)$.

2. **Rewrite $n$ in the desired form**:
We need to express $n$ as $n = ab + bc + ca$. This can be rewritten as:
$$n = c(a + b) + ab$$
This implies that $a + b$ must divide $n - ab$.

3. **Choose $a = k$ and $b = p - k$**:
Let $a = k$ and $b = p - k$. Then, we have:
$$a + b = k + (p - k) = p$$
Since $p \mid (n + k^2)$, we can write:
$$n + k^2 = mp \quad \text{for some integer } m$$
Therefore:
$$n = mp - k^2$$

4. **Determine $c$**:
We need to find $c$ such that:
$$c = \frac{n - k(p - k)}{p}$$
Substituting $n = mp - k^2$ into the equation, we get:
$$c = \frac{mp - k^2 - k(p - k)}{p} = \frac{mp - k^2 - kp + k^2}{p} = \frac{mp - kp}{p} = m - k$$
Thus, $c = m - k$.

5. Verify distinctness and positivity:
We need to ensure that $a$, $b$, and $c$ are distinct and positive integers.
- $a = k$
- $b = p - k$
- $c = m - k$

Since $k < p$, $a$ and $b$ are distinct. Also, since $p \leq 2\sqrt{n}$, $m$ is an integer such that $m \geq 1$. Therefore, $c = m - k$ is positive if $m > k$.

6. Check for contradictions:
- If $a = c$, then $k = m - k$, which implies $2k = m$. This would mean $2k \leq 2\sqrt{n}$, but since $n$ is square-free and even, this leads to a contradiction.
- If $b = c$, then $p - k = m - k$, which implies $p = m$. This would mean $p \leq 2\sqrt{n}$, but since $n$ is square-free and even, this also leads to a contradiction.

Therefore, the given construction works, and we have found pairwise distinct positive integers $a, b, c$ such that $n = ab + bc + ca$.

$\blacksquare$