Recall Euler’s Identity:
$$
\begin{align}
p(n \mid \text{ odd parts }) = p(n \mid \text{ distinct parts})
\end{align}
$$
Now consider the following exercises from 2.3 which is a refinement of Euler’s identity with an additional restriction
[2.3] Exercise 4: Euler's Identity with a Restriction
$$
\begin{align}
p(n \mid \text{ even number of odd parts}) = p(n \mid \text{ distinct parts, number of odd parts is even})
\end{align}
$$
That is, the number of integer partitions of \(n\) that must contain an even number of odd parts is the same as the number of integer partitions of \(n\) that must consist of distinct parts where the number of odd parts is even.
Example 1
Take \(n = 6\). The possible integer partitions of \(6\) are as follows:
$$
\begin{align}
6 &= 1 + 1 + 1 + 1 + 1 + 1 \\
6 &= 2 + 2 + 1 + 1 \\
6 &= 2 + 1 + 1 + 1 + 1 \\
6 &= 4 + 1 + 1 \\
6 &= 4 + 2 \\
6 &= 3 + 1 + 1 + 1 \\
6 &= 3 + 2 + 1 \\
6 &= 3 + 3 \\
6 &= 2 + 2 + 2 \\
6 &= 5 + 1 \\
6 &= 6
\end{align}
$$
The possible partitions with only odd parts are
$$
\begin{align}
6 &= 1 + 1 + 1 + 1 + 1 + 1 \\
6 &= 3 + 1 + 1 + 1 \\
6 &= 3 + 3 \\
6 &= 5 + 1 \\
\end{align}
$$
Observe here that all \(4\) partitions satisfy the condition that the number of odds are even. Now, if we take each one and merge equal parts repeatedly then for example
$$
\begin{align}
3 + 1 + 1 + 1 &\rightarrow 3 + 2 + 1
\end{align}
$$
Another example is
$$
\begin{align}
1 + 1 + 1 + 1 + 1 + 1 &\rightarrow 2 + 2 + 2 \rightarrow 4 + 2
\end{align}
$$
In fact, the resulting partitions are partitions of \(6\) into distinct parts, and each of them has an even number of odd parts as follows:
$$
\begin{align}
6 &= 6 \\
6 &= 4 + 2 \\
6 &= 3 + 2 + 1 \\
6 &= 5 + 1 \\
\end{align}
$$
Moreover, in each partition, we do have an even number of odds. This happens because whenever we merge two odd numbers, we reduce the number of odd numbers by two. Hence, the parity of the number of odd parts, remains the same.
References