# On questions involving the radical of an integer and different number theoretic functions: the Euler's totient function

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 $$\operatorname{rad}(n-\varphi(n))=\operatorname{rad}(n+\varphi(n)).\tag{2}$$

Question.

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.

• Upto $n=10^8$, every number $n$ satisfying $(2)$ also satisfies $rad(n)=6$ – Peter May 6 '18 at 14:06

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 $n=10^8$

• Many thanks for your attention and answer, I am going to study it. I am going to do the calculations, it's incredible! – user243301 May 6 '18 at 14:11

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 $p=2a+1=3$

• Many thanks for this new answer. I've read it a first time. – user243301 May 7 '18 at 11:55
• I hope it is correct. – Peter May 7 '18 at 11:56