First step with the equation $\sigma(n)=\phi(n)+\operatorname{rad}(n)$: a first statement

Question

I am wondering about the equation $$\sigma(n)=\phi(n)+\operatorname{rad}(n),\tag{1}$$ where for integers $n\geq 1$ we denote with $\sigma(n)$ the sum of divisors function, $\phi(n)$ is Euler's totient function and with $\operatorname{rad}(n)$ we denote the radical of the integer $n$, see in Wikipedia this definition. I am inspired in the equation by professor Iannucci, see the nice [1].

Currently (I've no implemented the function $\operatorname{rad}(n)$) I don't know if such equation $(1)$ was in the literature, and if it is possible deduce some interesting fact. I've dilucidated that the only integer less than 74 that satisifies $(1)$, is $n=2$. And I've searched in OEIS several strings like than these sigma(n)-phi(n), rad(n)... I am searched the equation in OEIS, in about 32 pages.

Question. If you can identify the sequence associated to $$\sigma(n)=\phi(n)+\operatorname{rad}(n)$$ and if it was in the literature, please refer it. Additionally, if you can compute more terms of such sequence it is good. I am interested about what should be the first step to study the specific equation $(1)$ by means of mathematical reasoning with the purpose to set some statement related with it (not so professional as papers in a journal, I say the first interesting statement about this equation). Many thanks.

I know that is very important have more terms of our sequence, also the size of each side in the equation, as calculated Iannucci in the first paragraph, but I have no the average value of $$\phi(n)+\operatorname{rad}(n).$$ Finally I believe that it should be very imporant the parity of each of our summands.

Then, imagine that a professor ask you about how set some mathematics about this equation, what is your reasoning and statement/conjecture? I hope that my question isn't too broad. I am asking about the first and more important fact about this equation. After that I have some answer I should choose an answer.

References:

[1] Douglas E. Iannucci, On the Equation $\sigma(n)=n+\phi(n)$, Journal of Integer Sequences, Vol. 20 (2017), article 17.6.2.

Also could be interesting check Hasler's sequence A228947, also from the On-line Encyclopedia of integer sequences.

Answer 1

For prime powers $n=p^k$ we can immediately see that the equation does not hold, because $$ \frac{p^{k+1}-1}{p-1}=\sigma(n) \neq \phi(n)+\operatorname{rad}(n)=p^k-p^{k-1}+p. $$ Indeed, the left hand side is congruent $1$ modulo $p$. Next, try to consider the case where $n$ is divisible by $2$ primes.

Answer 2

I wrote some Python code and checked up to 1,000,000 and found nothing except $n=2$ that satisfied the equation. In this range, there were near misses for the following values of $n$:

$n=1$: $\mbox{rad}(1)+\phi(1)=2$, $\sigma(1)=1$ (miss by $1$).

$n=3$: $\mbox{rad}(3)+\phi(3)=5$, $\sigma(3)=4$ (miss by $1$).

$n=15$: $\mbox{rad}(15)+\phi(15)=23$, $\sigma(15)=24$ (miss by $1$).

$n=21$: $\mbox{rad}(21) +\phi(21) = 29$, $\sigma(21) = 28$ (miss by $1$).

$n=903$: $\mbox{rad}(903)+\sigma(903)=1407$, $\sigma(903)=1408$ (miss by $1$).

Up to $1000000$, the above are all the values of $n$ that miss by $1$. There are no values that miss by $2$. The following values miss by $3$: $4, 5, 429, 861, 987$.

There are lots of values that miss by $4$ because for any odd prime $p$, we have $\mbox{rad}(2p)+\phi(2p)-\sigma(2p)=2p+(p-1)-(3(p+1))=-4$.

Answer 3

Using results from the Iannucci paper referenced in the problem statement, we can show that $n=2$ is the only solution to the problem. (Note: Eq (1) refers to the equation so-numbered above in the problem statement.)

Claim: Any solution to Eq (1) is square-free.

Proof: Suppose $n$ is not square-free. Say $n=p^k\cdot b$ for some prime $p$, with $k\ge 2$ and $p\nmid b$. We will show that $n$ is not a solution to Eq (1), and thus establish the Claim.

We have $$\begin{array}{cccr} \mbox{rad}(n)+\phi(n)&=&p\cdot\mbox{rad}(b)+(p^k-p^{k-1})\phi(b) \\ &\le& p\cdot b+(p^k-p^{k-1})b\\ &=& n+(p-p^{k-1})b& \\ &\le&n &(\mbox{because }k\ge 2) \\ &<&\sigma(n) \end{array}$$ Thus Eq(1) cannot hold for such $n$. So the Claim is established.

Now let $n$ be a solution of Eq(1). By the claim, $n$ is square-free. Hence $\mbox{rad}(n)=n$. And so in this situation, Eq (1) becomes $n+\phi(n)=\sigma(n)$.

But Iannucci shows (in Theorem 2 of the above referenced paper) that any integer $n\ge 2$ satisfying $n+\phi(n)=\sigma(n)$ must be an odd perfect square. But our candidate $n$ is square-free. We conclude that $n=2$ is the only solution to Eq (1).