How to find all even perfect number of a particular form.

Question

Find all the even perfect numbers of the form $a^a +1$, where $a \in \mathbb{N}$.

Can you please provide me some hint or idea on how to find such numbers.

Answer 1

The following notes are taken from the Introduction to a document sent to myself by Douglas Iannucci ("I" refers to "Iannucci"):

"At the $1990$ West Coast Number Theory Conference, John Selfridge posed a problem: to prove that $28$ is the only perfect number of the form $n^n + 1$. Peter Montgomery, in his usual fashion, had the problem solved before the conclusion of the conference. Selfridge and Montgomery then submitted the problem to the MAA Monthly, in which it appeared in $1992$ [2]. At that time I was a graduate student at Temple University. Luckily for me, Graeme Cohen was visiting Temple on a sabbatical, working with my dissertation advisor, Peter Hagis. I was able to prove that $28$ was the only even perfect number with the desired property, but I was stuck on trying to show that no odd perfect number had the form $n^n + 1$. I mentioned the problem to Graeme, who came up with a brilliant solution. I was happy when the Monthly printed our solution a couple of years later [1]." -- D. Iannucci

[1] G. Cohen and D. Iannucci, Solution to problem 10230, MAA Monthly, 1, (1995), 72.
[2] P. Montgomery and J. Selfridge, Proposal of problem 10230, MAA Monthly, 6, (1992), 570.

You can check out MAA Monthly problem $10230$ (and its solution) via JSTOR.

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Now, to address your specific question:

Since the perfect number $a^a + 1$ is even, it follows that $a$ is odd. But by the Euclid-Euler Theorem on the form of even perfect numbers, we have $$2^{p-1} (2^p - 1) = a^a + 1.$$ Since $a$ is odd, the RHS factors as $$a^a + 1 = (a + 1)(a^{a-1} - a^{a-2} + \ldots - a + 1) = cd,$$ where we set $$c = a+1$$ and $$d = a^{a-1} - a^{a-2} + \ldots - a + 1.$$ Note that $\gcd(c,d)=1$. From this, and the fact that $c < d$, it follows that $$2^{p-1} = c = a + 1$$ and hence, that $$2^p - 1 = 2c - 1 = 2(a + 1) - 1 = 2a + 1.$$

It follows that $$a^a + 1 = (a+1)(2a+1) = 2a^2 + 3a + 1,$$ an equation whose only solution (over the integers) is $a = 3$.

Hence, the only even perfect number of the form $a^a + 1$ is $3^3 + 1 = 28$.