Tales from the past

1. A Mail to Peter Scholze

While I finished my Abitur (the German A-levels equivalent) I read in a book about a number theoretic problem which has been open since Greek antiquity. The question was the following: Call a number n quasi perfect if the sum of all its divisors excluding 1 and n itself equals n. Are there any quasi perfect numbers?
When I looked the problem up on the internet, I learned that, apparently, no progress had been made at all on this question. My Abitur didn’t take an awful lot of time and I was looking for something to occupy myself with. I was sure that I was going to study mathematics, but without any knowledge beyond the basics I learned in school, this question seemed as good as any to start.
To my great surprise I did get somewhere: I could prove that only a square can be quasi perfect. To me that was quite exciting; True, the argument wasn’t extremely complicated but given that quite possibly nobody had ever thought about this question, something new! After I finished my A-levels, I wrote up my argument in a Latex document and sent it to one of the few mathematician I knew by name, Peter Scholze. I had read his name in a newspaper when he was appointed the youngest ever professor in Bonn, aged 24.
I had never written up a proof, so it turned out to be a bit, well, unconventional. I started my paper by arguing that even though it has been mathematical standard for 2000 years to count 1 as a divisor, one really ought not to do that and that for me, talking of a divisor of a number will exclude 1. I used my own definitions for everything and proved a bunch of, as I know now, trivialities and well known results. Among other things, I spent one paragraph proving that two (finite) sums are equal if both comprise the same summands. When I finished the paper, I had written up my short argument on 14 pages.
So imagine you are a rising star in mathematics, have just been appointed professor at age 24 and get a mail like that by somebody claiming to have proven something new about a 2000 year old problem. Would you reply? Would you even open the 14 pages document attached?Remarkably, Peter Scholze did. A few hours later, he sent me a long mail. He pointed me to references to some known things I had rediscovered, expressed my idea in a modern language and even spent a paragraph explaining the reasons why 1 is usually counted as a divisor. Or in his words:

“The problem was unknown to me, but the fact that only squares can be
quasi perfect is itself on the level of an exercise in number theory – with
which I don’t mean to take anything from your achievement, given that
you discovered all the basics yourself!”…”I was happy about your
argument modulo 3”… “I hope that you keep on enjoying mathematics,
perhaps even studying it soon at university. Your elaborations were
presented very clearly, even though the notation took me some getting used
to. Best wishes,
Peter Scholze”

For me, aged 17, this mail was incredibly encouraging. I never had any doubts about studying mathematics; but if I had had any, this mail surely would have dispelled them. Scholze’s mail was as nice as it could have been.Here is my simple proof that only squares can be quasi perfect in a modern language:

Assume that n=\prod_{i=1}^m p_i^{e_i} is a quasi perfect number.
Let as usual d(n) denote all divisors of n, including 1 and
n itself. I started by expressing

    \begin{equation*}2n+1=\sum_{d|n} d=\prod_i (1+p_i+ \dots + p_i^{e_i}).\end{equation*}

Part of my work was to rediscover the well-known identity in the second equality. The left hand side is odd, which means that all 1+p_i+ \dots + p_i^{e_i} must be odd. For p_i \neq 2 this means that e_i must be even. In particular, if n is not a square, n/2 is. Note that for any natural number a, a^2+1 is not divisible by three. Indeed, one of a-1, a, a+1 is divisible by 3, which implies that one of a^2-1=(a-1)(a+1) and a^2 is as well. This prevents a^2+1 from being divisible by three.

Thus, assuming that n is not a square, 2n+1=4 \cdot n/2+1 is not divisible by three. For p_j=2, 1+p^j+ \dots+p_j^{e_j} = 2^{e_j+1}-1. But 2^{2a}-1=(2^2-1)(1+ \dots+2^a) is divisible by three for any natural number a. If e_j were odd, this would yield a contradiction.
Thus, as all e_i are even, n is a square.

As I found out when writing this post, this fact was, perhaps unsurprisingly, already
Paolo Cattaneo (1951), ‚Sui numeri quasiperfetti‘, Boll. Un. Mat. Ital. (3) 6,

You can find Peter Scholze’s reply (in German) here.

The book mentioned above was: Underwood Dudley, „Mathematik zwischen Wahn und Witz.“