Are there infinitely many composite $a$ such that $\sum_{k=1}^{a}(k,a)\equiv1\pmod{a-1}$?

Question

Denote $\gcd(a, b)$ as $(a, b)$

Let $$f(a) =\sum_{k=1}^a (k,a)$$

Clearly $f(a)\equiv 1\pmod{a-1}$ if $a$ is prime.

Can it be shown that, there are infinitely many composite numbers satisfied $f(a)\equiv1\pmod{a-1}$?

First four composite numbers are $41124, 230867, 358267, 37539572$.

Formula

$$f(a)=\sum_{d\mid a} d\phi(a/d)$$

Thus $f$ is multiplicative.

We can compute $f(p^n)=np^{n-1}(p-1)+p^n$, when $p$ is prime. Which is a general formula for $f(a)$ in terms of the prime factorization of $a.$

Source Code PARI/GP:

forcomposite(a=2,10^8,s=0;fordiv(a,d,s+=d*eulerphi(a/d));if(s%(a-1)==1,print([a])))

Credit: viper found above four composite number through given code and Thomas Andrews, write a formula for $f(a)$.

This problem is sequel to my MSE post (link).Your suggestions, comments, the answer are very valuable to me. Apologies if the problem is just unsolvable. Thank you.

Answer 1

I do not know answer to your question, but here is an approach how one can extend a given integer $m$ to a solution $mp$ (if one exists) with a prime $p\nmid m$.

Using the multiplicativity, we want $$f(mp) = f(m)(2p-1) \equiv 1\pmod{pm-1}.$$ That is, there should be an integer $k$ such that $$f(m)(2p-1) - 1 = k(pm-1),$$ which is equivalent to $$(mp-1)(2f(m)-mk) = (m-2)f(m) + m.$$ Hence, we need to find a divisor $d$ of $(m-2)f(m) + m$ such that $d\equiv -1\pmod{m}$ and test if $p:=\frac{d+1}m$ is prime, and also if $p\nmid m$. From each such divisor, we obtain a suitable prime $p$.

Using the above approach for $m\leq 10^8$, I computed seven more solutions (although they may be not in order): $$148025049, 235167249, 242788284, 1085464188, 142772845653, 202728626748, 62763888399737.$$

Also, it can be used to come up with an heuristic argument (how likely for such $d$ to exist, how likely $\frac{d+1}m$ is prime etc.) for infinitude of solutions.