We denote for an integer $n>1$ its square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ with the definition $\operatorname{rad}(1)=1$. You can see this definition and the properties of this arithmetic function for example from this Wikipedia. In this post also we denote the Euler's totient function as $\varphi(n)$.

Today I am trying to explore claims involving the product of distinct primes dividing $n$, that is $\operatorname{rad}(n)$ and number theoretic functions. In this post we study the case about the Euler's totient function.

Claim 1. It is easy to prove that if $n$ has the form $n=2^{\alpha}5^{\beta}$, where $\alpha\geq 1$ and $\beta\geq 1$ are integers, then our integer $n$ satisfies $$\operatorname{rad}(2n+\varphi(n))=\operatorname{rad}(n+2\varphi(n)).\tag{1}$$

Claim 2. It is easy to prove that if $n$ has the form $n=2^{\alpha}3^{\beta}$, where $\alpha\geq 1$ and $\beta\geq 1$ are integers, then our integer $n$ satisfies $$\operatorname{rad}(n-\varphi(n))=\operatorname{rad}(n+\varphi(n)).\tag{2}$$


A) The first solution $m$ of $(1)$ that has a prime factor $P\notin\{2,5\}$ is $m=22110=2\cdot3\cdot 5\cdot 11\cdot 67$. Are there infinitelty many solutions $m$ of $(1)$ such that $m\notin$A033846 from the OEIS (these are integers of the form $n=2^{\alpha}5^{\beta}$, for some integers $\alpha\geq 1$ and $\beta\geq 1$)?

B) Prove or refute that if $n$ is an integer satisfying $(2)$ then $n$ has the form $n=2^{\alpha}3^{\beta}$, where $\alpha\geq 1$ and $\beta\geq 1$ are integers (that is the sequence A033845 from the OEIS).

Many thanks.

If isn't possible a full answer add what work can be done, I say it because I think that this kind of questions are very difficult to solve. Also I would like to add that I don't know if the equations $(1)$ or $(2)$ are in the literature. This is the link for OEIS.

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    $\begingroup$ Upto $n=10^8$, every number $n$ satisfying $(2)$ also satisfies $rad(n)=6$ $\endgroup$
    – Peter
    May 6, 2018 at 14:06

2 Answers 2


The proof for the second part is very long :

To show that $(2)$ can only be satisfied by $rad(n)=6$ , first of all note that for a positive integer $n$ and $r:=rad(n)$, we have $$\alpha:=\frac{\phi(n)}{n}=\frac{\phi(r)}{r}$$ because the prime factors of $n$ and $r$ coincide.

Hence, we can assume that $n$ is squarefree and we have to show $n=6$.

With this notation, $(2)$ turns into $$rad(n-\alpha n)=rad(n+\alpha n)$$

This cannot hold if $n-\alpha n$ and $n+\alpha n$ do not share the same prime factors. We assume $n>1$ to avoid the issue to deinfe $rad(0)$.

Now , suppose $\alpha:=\frac{a}{b}$. Because of $\phi(n)<n$, we can assume $0<a<b$ and $\gcd(a,b)=1$

Now we have $$n-\alpha n=\frac{b-a}{b}\cdot n$$ and $$n+\alpha n=\frac{b+a}{b}\cdot n$$ It is clear that neither $b-a$ not $b+a$ shares a prime factor with $b$ because then, it would follow $p|a$ which contradicts $\gcd(a,b)=1$

If $p|b-a$ and $p|b+a$ , then we have $p|2b$ and $p|2a$. Since $p|a$ and $p|b$ cannot hold both, we can conclude $p=2$. So, $b-a$ and $b+a$ cannot share an odd prime factor.

So, the radicals can only coincide when $b-a$ and $b+a$ are both powers of $2$. If $a\ge 2$, $a$ and $b$ would both be even , which contradicts $\gcd(a,b)=1$. Hence $a=1$ , so $b-a$ and $b+a$ are both odd.

This rules out more than one odd prime factor of $n$ because then $\phi(n)$ would be divisible by $4$ , but not $n$, so $a$ would be even. The case $n=p$ gives $\phi(n)=p-1$ and forces $a=p-1$ and $b=p$ , so $a=1,b=2$ , but $b+a=3$ and $b-a=1$ do not share prime factor $3$

In the case $n=2p$ , $p$ an odd prime , we have $\phi(n)=p-1$ and we must have $a=\frac{p-1}{2}$ , $b=p$ , so we get $p=2a+1=3$

  • $\begingroup$ Many thanks for this new answer. I've read it a first time. $\endgroup$
    – user243301
    May 7, 2018 at 11:55
  • $\begingroup$ I hope it is correct. $\endgroup$
    – Peter
    May 7, 2018 at 11:56

Partial answer :

Suppose $$n=2^a\cdot 3^b\cdot 5^c\cdot 11^d\cdot 67^e$$ with positive integers $a,b,c,d,e$. It is easy to show that then $\phi(n)=\frac{16}{67}\cdot n$

So, we have $$2n+\phi(n)=\frac{150}{67}\cdot n$$ and $$n+2\phi(n)=\frac{99}{67}\cdot n$$ So, both sides have the prime factors $2,3,5,11$ and both have $67$ or not as well depending on $e$ and no other prime factors exist, hence the radicals coincide. It seems that only those numbers and $2^a\cdot 5^b$ do the job. I did not find an example yet, where the radical of $n$ was different from $10$ and $2\cdot 3\cdot 5\cdot 11\cdot 67$. There is no counterexample upto $n=10^8$

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    $\begingroup$ Many thanks for your attention and answer, I am going to study it. I am going to do the calculations, it's incredible! $\endgroup$
    – user243301
    May 6, 2018 at 14:11

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