Let $p$ be a prime number. Find all integers $k$ for which $\sqrt{k^2 -pk}$ is a positive integer.
1. We start with the equation \(\sqrt{k^2 - pk} = n\), where \(n\) is a positive integer. Squaring both sides, we get:
   \[
   k^2 - pk = n^2
   \]
   Rearranging, we have:
   \[
   k^2 - pk - n^2 = 0
   \]

2. We need \(k^2 - pk\) to be a perfect square. Let's denote \(k^2 - pk = m^2\) for some integer \(m\). Thus, we have:
   \[
   k^2 - pk = m^2
   \]
   Rearranging, we get:
   \[
   k^2 - pk - m^2 = 0
   \]

3. We now consider the factorization of the quadratic equation:
   \[
   k^2 - pk - m^2 = 0
   \]
   This can be factored as:
   \[
   (k - \frac{p}{2})^2 - (\frac{p^2}{4} + m^2) = 0
   \]
   However, this approach does not simplify the problem directly. Instead, we will analyze the divisibility properties.

4. **Claim:** \(p \nmid k\). **Proof:** Suppose \(p \mid k\). Then let \(k = pn\). Substituting \(k = pn\) into the equation, we get:
   \[
   (pn)^2 - p(pn) = m^2
   \]
   Simplifying, we have:
   \[
   p^2n^2 - p^2n = m^2
   \]
   Dividing both sides by \(p^2\), we get:
   \[
   n^2 - n = \left(\frac{m}{p}\right)^2
   \]
   Since \(n^2 - n\) is an integer, \(\left(\frac{m}{p}\right)^2\) must also be an integer, implying \(p \mid m\). Let \(m = pq\). Then:
   \[
   n^2 - n = q^2
   \]
   However, \(n^2 - n = q^2\) implies \(q^2 < n^2\), which is impossible for all \(n > 1\). Therefore, \(p \nmid k\). \(\blacksquare\)

5. Since \(\gcd(p, k) = 1\), we also have \(\gcd(k - p, k) = 1\). Therefore, \(k\) and \(k - p\) must both be squares or both be the negations of squares.

6. If both \(k\) and \(k - p\) are positive squares, let \(k = m^2\). Then:
   \[
   m^2 - p = x^2
   \]
   for some integer \(x\). This can be rewritten as:
   \[
   m^2 - x^2 = p
   \]
   Factoring, we get:
   \[
   (m - x)(m + x) = p
   \]
   Since \(p\) is a prime, the factors must be \((m - x, m + x) = (1, p)\) or \((p, 1)\). Solving these, we get:
   \[
   m - x = 1 \quad \text{and} \quad m + x = p
   \]
   Adding these equations, we get:
   \[
   2m = p + 1 \implies m = \frac{p + 1}{2}
   \]
   Subtracting these equations, we get:
   \[
   2x = p - 1 \implies x = \frac{p - 1}{2}
   \]
   Therefore, \(k = m^2 = \left(\frac{p + 1}{2}\right)^2\).

7. If both \(k\) and \(k - p\) are the negations of squares, let \(k = -m^2\). Then:
   \[
   -m^2 - p = -x^2 \implies m^2 = x^2 + p
   \]
   This implies:
   \[
   m^2 - x^2 = p
   \]
   which is the same equation as before, leading to the same solutions.

8. For \(p = 2\), the equation becomes:
   \[
   k^2 - 2k = n^2
   \]
   which does not yield any integer solutions for \(k\).

Conclusion:
The desired solutions are:
\[
k = \left(\frac{p + 1}{2}\right)^2 \quad \text{and} \quad k = -\left(\frac{p - 1}{2}\right)^2
\]
for odd primes \(p\), and no solutions exist for \(p = 2\).

The final answer is \( \boxed{ k = \left(\frac{p + 1}{2}\right)^2 \quad \text{and} \quad k = -\left(\frac{p - 1}{2}\right)^2 } \) for odd primes \(p\), and no solutions exist for \(p = 2\).