All polynomials with no natural roots and integer coefficients such that ϕ(n)|ϕ(P(n))\phi(n)|\phi(P(n))

Let P(x) be a polynomial with integer coefficients such that the equation P(x)=0 has no positive integer solutions. Find all polynomials P(x) such that for all positive integers n we have ϕ(n)ϕ(P(n)). It is conjectured there are none (other than the trivial P(x)=xkQ(x)).

NOTE: For ϕ(P(n)) to be well-defined, it has been suggested that we require P(n)>0 for all positive integers n.

Answer

As this question has not been answered for a very long time, I think it is appropriate to use a strong conjecture to resolve it, namely Schinzel’s hypothesis H. Let us write P(x)=P1(x)k1...Pr(x)kr where Pi(x)Z[x] are irreducible, distinct, non-constant, and have positive leading coefficient (possibly r=1). Let the constant coefficient of Pi be ci. We may assume that ci are all non-zero, since otherwise we arrive at the trivial case xP(x). We have
ϕ(n)ϕ(P(n))=ϕ(P1(n)k1...Pr(n)kr)
for all n1. Let c=|c1...cr|, and define the polynomials Qi(x)=Pi(cx)|ci|Z[x]. Then we have
ϕ(cn)ϕ(Q1(n)k1...Qr(n)kr|c1|k1...|cr|kr).
for all n1. The advantage of this was that Qi have their constant coefficients equal to ±1. Let D be large enough, in particular D>degP (but we impose another condition later). We have in particular
ϕ(cD!n)ϕ(Q1(D!n)k1...Qr(D!n)kr|c1|k1...|cr|kr).
for all n, and these new polynomials have no small prime divisors. Finally, we apply Schinzel’s hypotheisis H to the irreducible polynomials x,Q1(D!x),...,Qr(D!x) (by Gauss’ lemma, f(x) is irreducible in Z[x] if and only if f(kx) is). Their product does not have any fixed prime divisor q because then we would have
\begin{align*}
\prod_{i=1}^r Q_i(D!x)\equiv 0 \mod q
\end{align*}

for 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 that
\begin{align*}
p-1\mid 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*}

Since Q_i(D!p)\equiv Q_i(D!)\pmod{p-1}, we get
\begin{align*}
p-1\mid 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*}

However, we can choose 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 large enough.

I doubt that this problem can be solved without assuming some conjecture as \phi(n) is surprisingly hard to control for general n; for example, Lehmer’s totient problem about solving \phi(n)\mid n-1 seems perhaps easier but is known to be open.

So, to gain some control, one would like to choose 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, although most numbers are squarefree and any polynomial with no fixed prime divisor should work. There are results about polynomials having infinitely almost prime values, but I am not sure whether they would be very helpful here.

Attribution
Source : Link , Question Author : Amir Parvardi , Answer Author : This site has become a dump.

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