TITLE: Has this strong number theoretic conjecture of Euler been proved, and where could I find such a proof?
QUESTION [16 upvotes]: Polya cites this work of Euler as an example of a conjecture which Euler considered impossible to doubt, and yet still needing a demonstration.
It is on pages 90-98 of "Induction and Analogy in Mathematics", which is volume I of "Mathematics and Plausible Reasoning", Princeton 1954. The reference to Euler is in a footnote on page 90.  (Opera Omnia ser 1 vol 2 p 241-253).
The conjecture is a recursion formula for the sum of the divisors of any integer, it is stated on page 93 of Polya.
Thanks for any help. 
Edit: In response to comments, here is the conjecture. For $n \geq 1$, let $\sigma(n) = \sum_{d|n} d$ be the sum of divisors of $n$. Euler conjectures the remarkable formula 
$$\begin{align*}
&\ \sigma(n)\\ 
=&\ \sigma(n-1) + \sigma(n-2) - \sigma(n-5) - \sigma(n-7) + \sigma(n-12) + \sigma(n-15) - \sigma(n-22) + \ldots \\
=&\ \sum_{k \geq 1} (-1)^{\text{floor}\left(\frac{k-1}{2}\right)} \sigma(n - f(k))
\end{align*}
$$ 
where $f(k)$ is defined recursively by $f(1) = 1$ and $f(k) = f(k-1) + \frac{k}{2}$ if $k$ is even, $f(k) = f(k-1) + k$ if $k$ is odd, for all $k > 1$. (Euler stipulates that $\sigma(n - f(k)) := 0$ if $n - f(k) < 0$, and one should substitute $n$ for $\sigma(0)$ if such a summand should arise on the right.) 
My question is whether this has ever been proved, and if so where I could find the proof.

REPLY [21 votes]: You can find the outline of a proof of this recursion relation in Surprising Connections between Partitions and Divisors. The line of reasoning first derives the recursion relation for the number of partitions $p(n)$ of $n$ from Euler's pentagonal number theorem, and then shows that the sum of divisors $\sigma(n)$ has the same recursion relation as $p(n)$.