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.
