### Instruction: For all composite integers $n$, what is the largest integer that always divides into the difference between $n$ and the cube of $n$?### Response: This problem is asking me to find the greatest common divisor (GCD) of $n - n^3$ for all composite $n$.
I can rewrite $n - n^3$ as $n(1 - n^2)$ using the distributive property.
If $n$ is composite, then it has at least one prime factor, say $p$.
Then $p$ divides both $n$ and $n(1 - n^2)$, so $p$ is a common divisor of $n - n^3$.
Is $p$ the largest possible common divisor?
If $n$ has another prime factor, say $q$, then $q$ also divides $n$ and $n(1 - n^2)$.
So $pq$ is also a common divisor of $n - n^3$.
In fact, any product of distinct prime factors of $n$ is a common divisor of $n - n^3$.
The largest such product is the same as $n$ itself.
So $n$ is the largest integer that always divides into the difference between $n$ and the cube of $n$.