About the proof of Euler’s Pentagonal Number Theorem on Wiki Euler’s Pentagonal Number Theorem on Wikipedia 
For convenience, here below is the statement:

Let $n$ be a nonnegative integer, let $q_e(n)$ be the number of partitions of $n$ into even number of distinct parts and $q_o(n)$ - number of partitions of $n$ into odd number of distinct parts. Then $q_e(n) - q_o(n) = \cases{(-1)^k \ \text{ if } n = \frac{k(3k \pm 1)}{2} \\ 0  \text{  otherwise}}$.

I understand the proof all the way up to the point where it says "...in which case there is exactly one Ferrers diagram left over".
According to the proof, integer $12$ has precisely one single non-invertible partition which is $(5, 4, 3)$ and it contributes $(-1)^3$ to the coefficient of $x^{12}.$ Since $12$ is a relatively small integer, the claim could be checked manually (which I admit I hadn't done).
But how do we know the claim in bold holds for horrendously large numbers or in general? Exactly what part of the proof shows the claim in bold? Thanks.
 A: In the notation of that proof, the only non-invertible partitions are those with $m=s$ and those with $m=s+1$, and it’s clear that in each case the value of $m$ uniquely determines a single Ferrers diagram of each type and hence a unique $n$ and partition of that $n$. The only possible problem, then is that there might be some $n$ that had a single non-invertible partition of each type.
This, however, is not actually possible. If $n$ has a partition of the first type, then $n=k^2+\binom{k}2$, where $k$ is the number of parts (i.e., the $s$ of the proof). If $n$ also has a partition of the second type, then $n=\ell^2+\binom{\ell+1}2$, where $\ell$ is again the number of parts. Thus,
$$k^2+\binom{k}2=\ell^2+\binom{\ell+1}2\,,$$
so
$$\frac{k(3k-1)}2=\frac{\ell(3\ell+1)}2\,,$$
and hence $3k^2-k=3\ell^2+\ell$. But then $3(k-\ell)(k+\ell)=\ell+k$, and $k+\ell\ne 0$, so $3(k-\ell)=1$, which is impossible. Thus, $n$ cannot have partitions of both types and therefore can have at most one non-invertible partition.
