0
$\begingroup$

For integers $n\geq 1$ with $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ we denote the squarefree kernel or radical of an integer $n$ (see if you want this Wikipedia). And $\varphi(n)$ denotes the Euler's totient funciton. Then while I was stuying the equation $$\operatorname{rad}(\varphi(2\cdot n))=2$$ in the context of constructible polygons with compass and straightedge, see this section of this Wikipedia, I wondered next variation $$\operatorname{rad}(\varphi(3\cdot n))=2\cdot 3.\tag{1}$$

Ater I search the resulting sequence of solutions of our equation $(1)$, the first few terms are $3,6,7,9,12,13,14,\ldots$, in The On-Line Encyclopedia of Integer Sequences, I would like to do a comparison to the sequence A135412. I know that there are terms in A135412, but don't satisfy our equation (1). You can read the definition of Heronian mean from this Wikipedia.

Question. I would like to know* if it is possible to prove or refute that:

If $n$ satisfies $$\operatorname{rad}(\varphi(3 n))=6$$ then our solution $n$ equals three times the Heronian mean of two positive integers.

Many thanks.

I think that this question is curious. If a full answer isn't feasible I'm interested to know what work can be done.

*Additionally if you want to add remarks in your answer about different equations involving the Euler's totient function and/or the squarefree kernel, showing some remarkable relationship to the sequence of the Heronian means in some way feel free to provide me such feedback.

$\endgroup$
  • 1
    $\begingroup$ This gets somewhat interesting if you require the Heronian mean of two distinct positive integers. You might look at the appropriate quadratic forms. Gerhard "Too Easy As Currently Written" Paseman, 2018.02.27. $\endgroup$ – Gerhard Paseman Feb 27 '18 at 19:54
  • 1
    $\begingroup$ And indeed, primes dividing a^2 +ab +b^2 for a and b coprime are of the form 3 or 6n+1. So I imagine your result boils down to a representation by a certain quadratic form. Gerhard "Cyclotomic Polynomials To The Rescue" Paseman, 2018.02.27. $\endgroup$ – Gerhard Paseman Feb 27 '18 at 19:59
  • 1
    $\begingroup$ I proved your conjecture below. $\endgroup$ – GH from MO Feb 27 '18 at 21:31
2
$\begingroup$

Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of some Eulerian integer $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.

$\endgroup$
  • $\begingroup$ @user75829: For an odd prime $2n-1$, it is straightforward that $\operatorname{rad}(\varphi(2n-1))=\operatorname{rad}(n-1)$ holds if and only if $n$ is odd. So what you are saying is that for an odd composite number $2n-1$, the equation never holds. If this is true, then it is probably very difficult to prove. $\endgroup$ – GH from MO Mar 1 '18 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy