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

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.

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.

• If some user wants to study this equation, then he/she is welcome to do it. Thanks. – user243301 Jul 1 '17 at 0:59
• And I belive that could be interesting (I did some calculations and reasonings) the similar equation than $(9)$ of this MathWorld $$\sigma(n)+\phi(n)=n+\operatorname{rad}(n).$$ – user243301 Jul 2 '17 at 10:26

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.

• Many thanks, I am searching these kind of claims, as your statement. Thus a first step is to study what happens with prime powers. Perfect reasoning!!! – user243301 Jun 30 '17 at 15:02

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 $\mbox{rad}(2p)+\phi(2p)-\sigma(2p)=2p+(p-1)-(3(p+1))=-4$.

• Many thanks you are very generous with your answer. And your works is also professional. This site MSE is possible by the merit and effort as users as you or as professor Burde. And now your calculations are a good reference about this problem. – user243301 Jul 4 '17 at 15:55

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).

• Many thanks I am going to take notes in my notebook about your statement to study it. Many thanks for share your statement with us. – user243301 Jul 5 '17 at 16:47