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.

I know that there is literature about inequalities concerning particular values of the Euler's totient function $\varphi(m)$, say us comparisons of the type $$\varphi(an+b)>\varphi(cn+d).$$

Question. I'm curious about what can be interesting comparisons of particular values of the square-free kernel, say us $$\operatorname{rad}(an+b),$$ where $a,b$ are simple cases of integers, and the quantity $$\varphi(cn+d),$$ with $c,d$ also simple integers, by means of an inequality $>$ or well $<$. Many thanks

Thus I am asking about the inequalities that I've evoked: I am asking about an inequality/comparison $$\operatorname{rad}(an+b)<\varphi(cn+d)$$ or well $$\varphi(cn+d)<\operatorname{rad}(an+b),$$ for a simple choice of interegers $a,b,c$ and $d$, with a good mathematical content.

Of course if to make an asymptotic comparison for sufficiently large $n'$s you need to introduce in RHS terms like $$\text{the first arithmetic function}<\underbrace{A(n)\cdot\text{the second arithmetic function}+\text{error term}}_{\text{this RHS involves particular values of the second arithmetic function}},$$ being $A(n)$ a function (or a constant), feel free to do such comparison.

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    $\begingroup$ If $a = c = 1$ and $b = d = 0$, then both inequalities are false for infinitely many $n$; moreover, neither can be fixed by multiplying/dividing by a large constant. If $n = p^k$ for some prime $p$ and positive integer $k$, we have $\text{rad}(n) = p$ while $\phi(n) = p^k - p^{k-1}$. If $n = p_1 \cdots p_k$ for $p_1, \cdots, p_k$ distinct primes we have $\text{rad}(n) = n$ and $\phi(n) = n \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right)$, and the latter can be made arbitrarily small compared to $n$. $\endgroup$ – Stanley Yao Xiao May 20 '18 at 16:49

We can prove that for any real number $M>1$, the inequalities $M\mathop{\rm rad}(an+b)<\phi(cn+d)$ and $\mathop{\rm rad}(an+b)>M\phi(cn+d)$ are each satisfied for infinitely many integers $n$.

First, note that we may assume that both $\gcd(a,b)=1$ and $\gcd(c,d)=1$: dividing $an+b$ by $\gcd(a,b)$ changes the $\mathop{\rm rad}(an+b)$ by only a bounded amount, and similarly dividing $cn+d$ by $\gcd(c,d)$ changes $\phi(cn+d)$ by only a bounded amount.

Next, note that if $an+b=cn+d$ then the argument in Stanley Yao Xiao's comment on the OP can be easy adapted to handle this case as well. So we may assume that $an+b \ne cn+d$.

To force $\mathop{\rm rad}(an+b)<\phi(cn+d)$, choose a large prime $p$, and restrict $n$ to the residue class modulo $p^2$ for which $p^2$ divides $an+b$. There are infinitely many $n$ in this residue class modulo $p^2$ for which $c'n+d'$ is prime, by Dirichlet's theorem. One can check that this gives $M\mathop{\rm rad}(an+b)<\phi(cn+d)$ when $p$ is large enough.

To force $\phi(cn+d)<\mathop{\rm rad}(an+b)$, let $q$ be a highly composite number coprime with $abcd$, and choose a residue class modulo $q$ such that $q$ divides $cn+d$. (Perhaps it's necessary to choose $q$ squarefree and choose a residue class modulo $q^2$ such that $\gcd(q^2,cn+d)=q$.) One can show that there are infinitely many $n$ in this residue class such that $an+b$ is squarefree, and then check that $M\phi(cn+d)<\mathop{\rm rad}(an+b)$ for such $n$ when $\phi(q)/q$ is small enough.

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  • $\begingroup$ Many thanks to you, and again to Yao Xiao. I am going to study your reasoning. $\endgroup$ – user75829 May 20 '18 at 19:16

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