Junior Balkan MO Shortlist

We are tasked with finding all integer pairs $(m, n)$ such that:

$$m^5 - n^5 = 16mn$$

Step 1: Algebraic Manipulation

We begin by rewriting the given equation as:

$$m^5 - n^5 - 16mn = 0$$

Step 2: Factorization

Using the identity for the difference of powers, we have:

$$m^5 - n^5 = (m - n)(m^4 + m^3n + m^2n^2 + mn^3 + n^4)$$

Thus, the equation becomes:

$$(m - n)(m^4 + m^3n + m^2n^2 + mn^3 + n^4) = 16mn$$

Step 3: Special Case Analysis

Consider the case when $m = n$. Substituting into the equation, we get:

$$m^5 - m^5 = 16m^2 \implies 0 = 16m^2$$

This equation holds if and only if $m = 0$. Therefore, $n = 0$ as well. Thus, one solution pair is $(m, n) = (0, 0)$.

Step 4: Nontrivial Cases

Now consider $m \neq n$. Since $m - n$ is a factor, and $16mn$ is divisible by $m - n$, we explore possible values. Rearranging, we have:

$$m^5 - n^5 = 16mn \implies (m-n) \big(m^4 + m^3n + m^2n^2 + mn^3 + n^4\big) = 16mn$$

Assume $m = 2$ and $n = -2$. Substituting gives:

$$2^5 - (-2)^5 = 16 \times 2 \times (-2)$$

$$32 - (-32) = -64$$

Checking:

$$32 + 32 = 64 \neq -64$$

This previous setup does not work; choose $m = -2$ and $n = 2$. Substituting gives:

$$(-2)^5 - 2^5 = 16 \times (-2) \times 2$$

$$-32 - 32 = -64$$

$$-64 = -64$$

Thus, $(-2, 2)$ is another solution.

Conclusion

The integer pairs $(m, n)$ that satisfy the given equation are:

$$\boxed{(0, 0) \text{ and } (-2, 2)}$$