For which even positive integers $m, n$, and $p$ is it true that

$$
\sqrt[n]{m-p}=\sqrt[m]{n-p}
$$
The numbers under the square roots must be non-negative, since the root index on both sides is even (positive) integers. Thus, $m p \geq p$ and $n \geq p$. Raising both sides of the equality to the $(m \cdot n)$-th power, we arrive at the relationship

$$
(m-p)^{m}=(n-p)^{n}
$$

If $m>n \geq p$, then $m-p>n-p \geq 0$. Consequently, $m-p$ is at least two, since it is an even number. Then, because $m>n$,

$$
(m-p)^{m}>(n-p)^{n}
$$

since for a base greater than one, $a^{x}$ is strictly monotonically increasing. For non-negative $x$, $x^{n}$ is also monotonically increasing, so

$$
(m-p)^{n} \geq(n-p)^{n}
$$

We have found that for $m>n \geq p$,

$$
(m-p)^{m}>(n-p)^{n}
$$

thus (1) has no solution. Similarly, it can be shown that (1) does not hold for $n>m \geq p$ either. The case $m=n \geq p$ remains. It is obvious that in this case $\sqrt[n]{m-p}=\sqrt[m]{m-p}=\sqrt[m]{n-p}$.

Therefore, the equality in question is satisfied by the positive even numbers that meet the condition $m=n \geq p$, and only by these.