To prove these divisibility statements, let's go through them one by one.

### a) \( 15^n + 15^{n+2} \) chia hết cho 113 với mọi số tự nhiên \( n \).

Given expression: \( 15^n + 15^{n+2} = 15^n (1 + 15^2) \).

First, calculate \( 15^2 \mod 113 \):

\[
15^2 = 225
\]

Now, compute \( 225 \mod 113 \):

\[
225 \div 113 \approx 1 \quad \Rightarrow \quad 225 - 113 \times 1 = 112
\]

So \( 15^2 \equiv 112 \pmod{113} \).

Now, \( 1 + 15^2 \equiv 1 + 112 \equiv 113 \equiv 0 \pmod{113} \).

Thus, \( 15^n (1 + 15^2) \equiv 0 \pmod{113} \), meaning the expression is divisible by 113 for any natural number \( n \).

### b) \( n^4 - n^2 \) chia hết cho 4 với mọi số tự nhiên \( n \).

Expression: \( n^4 - n^2 = n^2(n^2 - 1) \).

The factor \( n^2(n^2 - 1) \) includes two consecutive numbers \( n^2 \) and \( n^2 - 1 \), so one of them is always even.

Furthermore, for any integer \( n \):

- If \( n \equiv 0 \pmod{2} \) or \( n \equiv 1 \pmod{2} \), both \( n^2 \equiv 0 \pmod{2} \), so \( n^2(n^2 - 1) \equiv 0 \pmod{4} \).

Therefore, \( n^4 - n^2 \equiv 0 \pmod{4} \) for any natural number \( n \).

### c) \( n^3 - n \) chia hết cho 6 với mọi số nguyên \( n \).

Expression: \( n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1) \).

The expression \( n(n - 1)(n + 1) \) includes three consecutive integers. Thus:

- At least one of these integers is divisible by 2.
- At least one is divisible by 3 (since every three consecutive integers include a multiple of 3).

Hence, their product is divisible by \( 2 \times 3 = 6 \).

Thus, \( n^3 - n \equiv 0 \pmod{6} \) for any integer \( n \).

In conclusion, all statements are proved to be true.