The integer values of $\sum_{d|n}\frac{\sigma(n/d)^d}{d}$.

Question

I was looking for integer values of the $$\sum_{d|n}\frac{\sigma(n/d)^d}{d}$$ where the $\sigma(n)$ is a divisors sum of $n$.

And amazingly has found that the only integer values for $n<1500000$ are: $$1, 39, 793$$

So I assume that these are the only integer numbers of this kind.

Any ideas how to prove or disprove this?

The according OEIS sequence is: https://oeis.org/A268983

EDITED: With @RobertIsrael help in question The solution of congruences system. and according to @user1952009 comment I have found the next item in this sequence: $$2408321608150261253061174553 = 22419767768701 * 107419560853453$$

Answer 1

Let $$f(n) =\sum_{d | n} \frac{\sigma(n/d)^d}{d}$$ If $p,q$ are two different primes then $$f(pq) =\sigma(pq) + \frac{p(p+1)^q + q (q+1)^p + 1}{pq}$$ For $f(pq) \in \mathbb{Z}$ we need $p(p+1)^q + q (q+1)^p + 1 \equiv 0 \bmod p$ and $q$

$$\implies \qquad q (q+1)\equiv -1 \bmod p, \qquad p (p+1)\equiv -1 \bmod q$$ (by the Fermat little theorem)

Letting $g(n) = \sum_{d | n} \frac{\sigma(n/d)}{d}$ which is multiplicative, then $f(pq) \in \mathbb{Z}$ iff $$\sigma(pq)+\frac{p(p+1) + q (q+1) + 1}{pq}=g(pq)=g(p)g(q)= (p+1+\frac{1}{p})(q+1+\frac{1}{q}) \quad \in \mathbb{Z}$$

---

With this matlab program I didn't find more solutions :

for a = [1:3200]
    A = a*(a+1)+1; fac = factor(A);
    for j = 1:length(fac)
        p = fac(j);
        if p > a
            P = p*(p+1)+1; fac2 = factor(P);
            for j2 = 1:length(fac2)
                q = fac2(j2);
                if mod(q,p) == a    fprintf('%d %d \n', p,q);   end
            end
        end
    end
end