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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 $\phi(n) \mid \phi(P(n))$. It is conjectured there are none (other than the trivial $P(x) = x^k Q(x)$).

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

6005
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Amir Parvardi
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    @Amir: I have a silly question. By natural roots of $P$, do you mean positive integer solutions to $P(n) = 0$? – JavaMan Jun 03 '11 at 14:03
  • @DJC: Yes, sorry if it's not clear. It would be very long if I wanted to write it completely. Thanks for mentioning. – Amir Parvardi Jun 03 '11 at 14:20
  • @Amir: (: I figured so but was just checking. – JavaMan Jun 03 '11 at 14:21
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    @Amir, better be long and clear. – lhf Jun 03 '11 at 14:30
  • @Amir: Do you know of _any_ polynomial which satisfies your condition? – JavaMan Jun 03 '11 at 15:46
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    @DJC: What about $p(x)=x$. – Eric Naslund Jun 03 '11 at 16:35
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    I suppose _positive solutions_ to $P(n) = 0$ should be changed to _nonnegative solutions_ to $P(n) = 0$. Are there _any_ nontrivial examples? – JavaMan Jun 03 '11 at 16:44
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    @Eric: The point of the problem is to find _all_ such polynomials. I only ask for one. If indeed there are no such polynomials, then these two tasks are one and the same. By the words "Classifying all such polynomials" I assumed that there might have been some nontrivial polynomials that fit the bill (other than those of the form $P(n) = n^kQ(n)$ which are excluded by the question). – JavaMan Jun 03 '11 at 18:25
  • @DJC: as far as I've seen "nice" polynomial problems, they all ask for finding "all" of such polynomials, however the only solution is the simplest and the most obvious one. :) – Amir Parvardi Jun 04 '11 at 02:41
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    Actually, this problem was posted [here](http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=406528). The first part have been solved, but the second part (this problem) is still unsolved. – Amir Parvardi Jun 04 '11 at 07:26
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    Shouldn't we actually require that $P(x) > 0$ for positive integers $x$, so that $\phi(P(x))$ is defined? For example, $P(x) = x^2-200$ is never zero but is negative for some values of $x$. – Srivatsan Aug 13 '11 at 03:35
  • Is it worth noting that the conjecture follows from the Germain prime conjecture? It says that there are infinitely many primes $p$ such that $2p+1$ is prime. – This site has become a dump. Jul 24 '13 at 10:36
  • @User84559, it doesn't since we have to show it for all polynomials, not just $P(x)=2x+1$. Another stab at it would be http://en.wikipedia.org/wiki/Bunyakovsky_conjecture but we would also require infinitely many prime values on _prime_ inputs. – Dan Brumleve Aug 10 '13 at 03:39
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    @Dan Brumleve, what I meant was that if $p>2$ and $q=2p+1$ are prime, then $2p\mid\phi(P(q))$ for such a polynomial, from which it follows that $q\mid P(q)$ and hence that the constant term of $P$ is divisible by $q$. – This site has become a dump. Aug 12 '13 at 15:18
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    I don't understand why the condition "nonconstant" is omitted in the statement of this problem. $p(x)=mx$ is a solution for every positive integer $m$. – Doc Oct 03 '13 at 17:57
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    $P(x)=x^n$ is an infinite family of solutions, unless "positive solutions" is changed to "nonnegative solutions" in the problem statement. (In other words, I wholeheartedly agree with JavaMan's comment above.) – Doc Oct 03 '13 at 23:13
  • The imprecise formulation of this problem is really bothering me. It should say: $P(x) > 0$ for all $x > 0$, and $P(0) \ne 0$ (thus excluding the trivial cases $P(x) = x^k Q(x)$, and making $\phi(P(n))$ well-defined). As it is currently stated, "it is conjectured that there are none" is out of place. Would anyone mind if I just changed it? – 6005 Jan 11 '14 at 09:32
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    2 Servaes: I do not see how you conclude that $q | P(q)$. In general, if $2p | \phi(m)$, it need not follow that $2p + 1$ divides $m$, e.g., $2*3 | \phi(13)$, but $7 \nmid 13$. – streetcar277 Feb 05 '14 at 13:02
  • Consider $n=2^k$. $\phi(2^k) = 2^{k-1}$, which grows exponentially. We require that $\phi(2^k) | \phi(P(2^k))$, so $\phi(P(2^k)) = 2^m$ for some $m < k$, which requires $P(2^k)$ is $2^m$ or $3 \times 2^m$ for some $m – Eric Towers Mar 21 '14 at 23:30
  • @EricTowers Except that the divisibility is supposed to work the other way round: $2^{k-1}$ should *divide* $\phi(P(2^k))$. – Peter Košinár Mar 28 '14 at 20:29
  • Is $\phi(n)$ the Euler function? – Beni Bogosel Apr 03 '14 at 14:03
  • Yes, it is, Beni. – Amir Parvardi Apr 04 '14 at 08:54

1 Answers1

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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 $P(x)=P_1(x)^{k_1}...P_r(x)^{k_r}$ 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 $x\mid P(x)$. We have \begin{align*} \phi(n)\mid \phi(P(n))= \phi(P_1(n)^{k_1}...P_r(n)^{k_r}) \end{align*} for all $n\geq 1.$ Let $c=|c_1...c_r|$, and define the polynomials $Q_i(x)=\frac{P_i(cx)}{|c_i|}\in \mathbb{Z}[x]$. Then we have \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*} for all $n\geq 1$. The advantage of this was that $Q_i$ have their constant coefficients equal to $\pm 1$. Let $D$ be large enough, in particular $D>\deg P$ (but we impose another condition later). We have in 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*} for all $n$, 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 $q$ 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.