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morpheus
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Como esta pregunta no ha sido respondida por un tiempo muy largo, creo que es adecuado el uso de un fuerte conjetura de resolver, a saber, Schinzel la hipótesis H. Vamos a escribir $P(x)=P_1(x)^{k_1}...P_r(x)^{k_r}$ where $P_i(x)\in \mathbb{Z}[x]$ are irreducible, distinct, non-constant, and have positive leading coefficient (possibly $r=1$). Let the constant coefficient of $P_i$ be $c_i$. We may assume that $c_i$ are all non-zero, since otherwise we arrive at the trivial case $x\mid P(x)$. Tenemos \begin{align*} \phi(n)\mid \phi(P(n))= \phi(P_1(n)^{k_1}...P_r(n)^{k_r}) \end{align*} para todos los $n\geq 1.$ Let $c=|c_1...c_r|$, and define the polynomials $Q_i(x)=\frac{P_i(cx)}{|c_i|}\in \mathbb{Z}[x]$. Entonces tenemos \begin{align*} \phi(cn)\mid \phi(Q_1(n)^{k_1}...Q_r(n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}). \end{align*} para todos los $n\geq 1$. The advantage of this was that $Q_i$ have their constant coefficients equal to $\pm 1$. Let $D$ be large enough, in particular $D>\deg P$ (pero imponer otra condición más adelante). Tenemos, en particular, \begin{align*} \phi(cD!n)\mid \phi(Q_1(D!n)^{k_1}...Q_r(D!n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}). \end{align*} para todos los $n$, and these new polynomials have no small prime divisors. Finally, we apply Schinzel's hypotheisis H to the irreducible polynomials $x,Q_1(D!x),...,Q_r(D!x)$ (by Gauss' lemma, $f(x)$ is irreducible in $\mathbb{Z}[x]$ if and only if $f(kx)$ is). Their product does not have any fixed prime divisor $q$ porque entonces tendríamos \begin{align*} \prod_{i=1}^r Q_i(D!x)\equiv 0 \mod q \end{align*} para $x=1,2,...,q-1$. However, this conguence has at most $\deg P$ solutions, so $q\leq \deg P+1$, which is a contradiction by the definition of $D$ (none of the polynomials $Q_i(D!x)$ is identically zero $\pmod q$ since their constant coefficients are $\pm 1$). Hence, we have infinitely many primes $p$ such that $Q_1(D!p),...,Q_r(D!p)$ are all primes. Setting $n=p$ in the previous formula involving $n$, we obtain by multiplicativity for large enough $p$ que \begin{align*} p-1\mediados de Q_1(D!p)^{k_1-1}(Q_1(D!p)-1)...(Q_r(D!p))^{k_r-1}(Q_r(D!p)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}). \end{align*} Desde $Q_i(D!p)\equiv Q_i(D!)\pmod{p-1}$, obtenemos \begin{align*} p-1\mediados de Q_i(D!)^{k_1-1}(Q_1(D!)-1)...Q_r(D!)^{k_r-1}(Q_r(D!)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}). \end{align*} Sin embargo, podemos optar $D$, depending only on the polynomial $P$, so that $Q_i(D!)\geq 2$ for all $i$, and then the right-hand side of the previous divisibility relation should be smaller than the left-hand side, which is a contadiction for $p$ lo suficientemente grande.
Dudo que este problema puede ser resuelto sin asumir algunas conjeturas como $\phi(n)$ is surprisingly hard to control for general $n$; for example, Lehmer's totient problem about solving $\phi(n)\mid n-1$ parece quizás más fácil, pero se sabe que estará abierto.
Así, para ganar un poco de control, a uno le gustaría elegir a $n$ in the relation $\phi(n)\mid \phi(P(n))$ to be prime, or at least closely related to them. However, very little is known about prime values of polynomials, and even less about their simultaneous prime values. In fact, we do not even know (according to this paper by M.Filazeta) a polynomial $f(x)\in \mathbb{Z}[x]$ such that $\deg f\geq 4$ and $f(x)$ is square-free for infinitely many integers $x$, aunque la mayoría de los números son squarefree y cualquier polinomio con no se ha fijado un divisor primo debería funcionar. No hay resultados acerca de los polinomios de tener infinitamente casi prime valores, pero no estoy seguro de si sería muy útil aquí.
