I am having some trouble with the following question.
Prove that for $n \ge 2$, $p(n) -p(n-2)$ is the number of partitions of $n$ with at most one part of size 1.
I can see that by differencing them, the two partitions is going to only produce partitions of size 1 and 2. What is the next step to prove that these partitions only contain at most one part of size 1?
For $n$ large enough there are $p(n-2)$ ways to partition integer $n$ is such a way that $1$ appears at least twice in it.
This because such a partition is exactly a partition of integer $n-2$ supplemented with two "ones".
There are $p(n)$ partitions of $n$ in total so $p(n)-p(n-2)$ of them do not have $1$ at least twice in it.