The following identity is presented an exercise:
[2.3] Exercise 3
Show that
$$
\begin{align}
p(n \mid \text{even parts}) = p(n/2) = p(n \mid \text{even number of each part})
\end{align}
$$
That is, the number of integer partitions made of even parts is the same as the number of integer partitions such that we have an even number of each part. These are also the same as the number of integer partitions of \(n/2\).
Example
Take \(n = 6\). Then using only even parts, we can write
$$
\begin{align}
6 &= 6 \\
6 &= 4 + 2 \\
6 &= 2 + 2 + 2
\end{align}
$$
For \(n/2 = 3\), we can partition it as
$$
\begin{align}
3 &= 3 \\
3 &= 2 + 1 \\
3 &= 1 + 1 + 1
\end{align}
$$
which correspond to partitions of \(6\) in which each part appears an even number of times
$$
\begin{align}
6 &= 3 + 3 \\
6 &= 1 + 1 + 1 + 1 + 1 + 1 \\
6 &= 2 + 2 + 1 + 1
\end{align}
$$
\(p(n \mid \text{even parts}) = p(n/2)\)
First note that if \(n\) is odd, then both sides are \(0\), since we can’t write an odd integer as a sum of even parts. Now, consider the following map: starting with even parts, divide each part by \(2\). Therefore
$$
\begin{align}
6 &= 6 \rightarrow 3 = 3 \\
6 &= 4+2 \rightarrow 3 = 2 + 1 \\
6 &= 2 + 2 + 2 \rightarrow 3 = 1 + 1 + 1
\end{align}
$$
The inverse of this map is multiplying everything by \(2\). Hence, this map is a bijection.
\(p(n \mid \text{even number of each part}) = p(n/2)\)
Consider the following map: starting with even number of each part, then for each pair of parts \(k+k\), replace it with a single \(k\). Then
$$
\begin{align}
6 &= 3 + 3 \rightarrow 3 = 3\\
6 &= 1 + 1 + 1 + 1 + 1 + 1 \rightarrow 3 = 1 + 1 + 1 \\
6 &= 2 + 2 + 1 + 1 \rightarrow 3 = 2 + 1
\end{align}
$$
The inverse of this map is just adding each part twice. Hence, this map is a bijection.
References