Prove that there are no self-conjugate partitions of n, whose third part is 2, when n is an odd integer.
My attempt: let n = 5, (1 + 1 + 1 + 1 + 1), (3 + 2), (3 + 1 + 1), (2+ 2 + 1), (4 + 1), (2 + 1 + 1 + 1). I see that the third part is odd (or 1), but how do I go about proving this?
Use the Ferrers diagram of a self-conjugate partition with third part 2 to show that $n=8+2k$ for some nonnegative integer $k$. You must have at least these 8: $$\begin{matrix} \bullet & \bullet & \bullet & \dots \\ \bullet & \bullet & \bullet & \dots \\ \bullet & \bullet & \circ & \circ \\ \dots & \dots & \circ & \circ \end{matrix}$$ And any additional ones appear in pairs because of self-conjugacy.