Let $p$ ($n|$distinct odd parts) be the number of partitions of $n$ into distinct odd parts. Prove that $p(n)$ is odd if and only if $p$($n|$distinct odd parts) is odd by using the theorem on self-conjugate partitions.
Ok, so we count the hooks in the Ferrers diagram, but how?
There may be an easier way, but the result follows easily from the one in this question, which has both a combinatorial answer and one that uses generating functions.
Partitions that aren't self-conjugate come in pairs, so $p(n)$ has the same parity as the number of self-conjugate partitions. But I assume that "the theorem on self-conjugate partitions" is the one that says the number of self-conjugate partitions is the number of partitions into distinct odd parts. So, $p(n)$ has the same parity as the number of partitions into distinct odd parts.
