What is the largest integer that must divide $n^5-5n^3+4n$ for all integers $n$?

[i]2016 CCA Math Bonanza Lightning #2.4[/i]
1. We start by factoring the expression \( n^5 - 5n^3 + 4n \):
   \[
   n^5 - 5n^3 + 4n = n(n^4 - 5n^2 + 4)
   \]
   Next, we factor \( n^4 - 5n^2 + 4 \):
   \[
   n^4 - 5n^2 + 4 = (n^2 - 4)(n^2 - 1)
   \]
   Further factoring \( n^2 - 4 \) and \( n^2 - 1 \):
   \[
   n^2 - 4 = (n - 2)(n + 2) \quad \text{and} \quad n^2 - 1 = (n - 1)(n + 1)
   \]
   Therefore, the expression can be written as:
   \[
   n(n^2 - 4)(n^2 - 1) = n(n - 2)(n + 2)(n - 1)(n + 1)
   \]

2. We need to determine the largest integer that must divide \( n(n - 2)(n + 2)(n - 1)(n + 1) \) for all integers \( n \). Consider the properties of these five consecutive integers:
   - Among any five consecutive integers, there must be at least one multiple of 3.
   - Among any five consecutive integers, there must be at least one multiple of 4.
   - Among any five consecutive integers, there must be at least one multiple of 5.

3. Additionally, since we have five consecutive integers, at least one of them must be even. However, since we have two even numbers among five consecutive integers, one of them must be a multiple of 4, and the other must be a multiple of 2. Therefore, the product must include \( 2 \times 4 = 8 \).

4. Combining these observations, the product \( n(n - 2)(n + 2)(n - 1)(n + 1) \) must be divisible by:
   \[
   \text{lcm}(3, 4, 5, 8) = 3 \times 5 \times 8 = 120
   \]

5. Therefore, the largest integer that must divide \( n^5 - 5n^3 + 4n \) for all integers \( n \) is \( 120 \).

The final answer is \( \boxed{120} \).