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We denote the Euler's totient function as $\varphi(n)$, and the radical of the integer $n> 1$ as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p,$$ taking $\operatorname{rad}(1)=1$ that is this definition from Wikipedia.

Then it is easy to prove the following claim (I don't know if this characterization of odd primes is in the literature).

Claim. An integer $n\geq 1$ is an odd prime if and only if satisfies $$\operatorname{rad}(2n)=n+\varphi(n)+1\tag{1}.$$

Proof. The difficult implication is by cases.$\square$

Then I wanted to use previous claim to get a characterization of twin primes. Because defining the arithmetic function $$f(n)=\frac{\operatorname{rad}(2n)}{n+\varphi(n)+1}\tag{2}$$ it is obvious that each twin prime pair $(n,n+2)$ solves the equation $$f(n)=f(n+2).\tag{3}$$

I would like to know what about the other implication as the next question.

Question. Prove or refute that if an integer $m\geq 1$ satisfies $(3)$, that is $$\frac{\operatorname{rad}(2m)}{m+\varphi(m)+1}=\frac{\operatorname{rad}(2m+4)}{m+\varphi(m+2)+3}$$ then $m$ and $m+2$ are primes (that is, thus $m$ is the lesser of a twin prime pair, sequence A001359 from the OEIS). Many thanks.

Feel free to add what work can be done, if you cann't provide us a full answer.

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Claim:$\;$If $n\in \mathbb{Z}^{+}$ is such that $f(n) = f(n+2)$, then $n$ is an odd prime.

Proof:

For convenience, the proof will be organized as a sequence of sub-claims, as listed below . . .

$\;\;\;(1)\;\,$If $p^2|n$, for some odd prime $p$, then $f(n) \ne f(n-2)$, and $f(n) \ne f(n+2)$.

$\;\;\;(2)\;\,$If $n$ is odd, but not prime, then $f(n) \ne f(n+2)$

$\;\;\;(3)\;\,$If $n$ is a multiple of $4$, then $f(n) < \frac{1}{2}$.

$\;\;\;(4)\;\,$If $n$ is even and squarefree, then $f(n) \ge \frac{1}{2}$.

$\;\;\;(5)\;\,$If $n$ is even, then $f(n) \ne f(n+2)$

Proof of $(1)$:

Suppose $p^2|n$, for some odd prime $p$.

For $k\in \mathbb{Z}^{+}$, let $a(k),b(k)$ be the numerator and denominator, respectively, of $f(k)\;$, when reduced to lowest terms.

From $p^2{\mid}n$, we get $p{\mid}\phi(n)$, hence $p{\not\mid}b(n)$.

It follows that $p{\mid}a(n)$.

From $p{\mid}n$, we get $p{\not\mid}(n-2)$, and $p{\not\mid}(n+2)$, hence $p{\not\mid}a(n-2)$, and $p{\not\mid}a(n+2)$.

It follows that $f(n) \ne f(n-2)$, and $f(n) \ne f(n+2)$.

This completes the proof of $(1)$.

Proof of $(2)$:

Suppose $n$ is odd, not prime, and $f(n) = f(n+2)$.

Note that $f(1) \ne f(3)$, hence $n > 1$.

Since $n$ is not prime, we have $\phi(n) < n-1$.

From $(1)$, it follows that $n$ and $n+2$ are squarefree, hence ${\text{rad}}(2n)=2n$, and ${\text{rad}}(2(n+2)) = 2(n+2)$.

Let $x=\phi(n)$, and let $y=\phi(n+2)$.

Then from $f(n) = f(n+2)$, we get $$ \frac{2n}{n+x + 1} = \frac{2n+4}{n+y+3} \qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\, $$ which yields $$ n = \frac{2(x+1)}{y-x} \qquad\qquad\qquad\qquad\; $$ It follows that $y > x$, hence, since $x,y$ are necessarily both even, we get $y-x \ge 2$. \begin{align*} \text{Then}\;\;n &= \frac{2(x+1)}{y-x}\\[4pt] &\le \frac{2(x+1)}{2}&&\text{[since $y-x \ge 2$]}\\[4pt] &= x+1\\[4pt] &=\phi(n)+1\\[4pt] &< (n-1) + 1&&\text{[since $\phi(n) < n-1$]}\\[4pt] &=n\\[4pt] \end{align*} contradiction.

This completes the proof of $(2)$.

Proof of $(3)$:

Suppose $n$ is a multiple of $4$.

Then from $2n = 8{\left({\large{\frac{n}{4}}}\right)}$, we get $\text{rad}(2n) \le {\large{\frac{n}{2}}}$, hence $$ f(n)=\frac{\text{rad}(2n)}{n+\phi(n)+1} < \frac{\left(\frac{n}{2}\right)}{n}=\frac{1}{2} $$ This completes the proof of $(3)$.

Proof of $(4)$:

Suppose $n$ is even and squarefree.

Then $\text{rad}(2n) = n$, hence $$ f(n)=\frac{\text{rad}(2n)}{n+\phi(n)+1} \ge \frac{n}{n+(n-1)+1} =\frac{1}{2} $$ This completes the proof of $(4)$.

Proof of $(5)$:

Suppose $n$ is even.

If at least one of $n,\,n+2$ is divisible by the square of an odd prime, then by $(1)$, we get $f(n) \ne f(n+2)$, and we're done.

Thus, we can assume neither of $n,\,n+2$ is divisible by the square of an odd prime.

It follows that one of $n,\,n+2$ is a multiple of $4$, and the other is even and squarefree, hence by $(3)$ and $(4)$, we get $f(n) \ne f(n+2)$, and again we're done.

This completes the proof of $(5)$.

From the truth of $(2)$ and $(5)$, the truth of the main claim follows.

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  • $\begingroup$ Many thanks tomorrow in the morning I am going to study your nice answer. $\endgroup$ – user243301 Mar 18 '18 at 21:45
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I think my previous answer leads to an immediate full resolution of your question.

You showed that $f(n)=1$ if and only if $n$ is an odd prime.

Thus, if $n$ and $n+2$ are twin primes, we get $$f(n)=f(n+2)=1$$ Conversely, suppose $f(n)=f(n+2)$, for some positive integer $n$.

Then from my previous answer, $n$ must be an odd prime.

Since $n$ is an odd prime, we get $f(n)=1$.

But then from $f(n)=f(n+2)$, we get $f(n+2)=1$, hence $n+2$ is an odd prime.

Therefore $n$ and $n+2$ are twin primes.

Thus, we have:

  • If $n$ is a positive integer, then $f(n)=f(n+2)$ if and only if $n$ and $n+2$ are twin primes.
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  • $\begingroup$ Many thanks for the clarification. $\endgroup$ – user243301 Mar 29 '18 at 17:03

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