9

Is there a polynomial, $f(x)$, such that for all natural numbers $p$ and $q$, if $\gcd(p, q) = 1$ then $\gcd(f(p), f(q)) = 1$?

Note : Function $f(x)$ must be a polynomial in $x$, not depend on $p$ or $q$, and not be the trivial case of a polynomial with only $1$ term ($f(x) = c$ or $f(x) = x^p$).

DanielV
  • 23,556
hanugm
  • 2,353
  • 1
  • 13
  • 34
  • At least $f(x) = x^2+x-1$ for $p=2$ and $q=3$, because $\gcd(5,11)=1$ – Eemil Wallin Jun 15 '15 at 09:57
  • It's not a polynomial, but if $F_n$ is the $n^{\text{th}}$ Fibonacci number, then $\gcd(p, q) = \gcd(F_p, F_q)$ : http://math.stackexchange.com/a/506108/97045 – DanielV Jun 15 '15 at 10:18
  • To clarify: do you want a polynomial that works for a specific pair of integers, or do you want a polynomial that works for all coprime integers? – Mathmo123 Jun 15 '15 at 10:18
  • I don't think so. –  Jun 15 '15 at 10:36
  • @Mathmo123 For all coprime integers. – hanugm Jun 15 '15 at 10:37
  • Your question is poorly worded. Do you mean that you are asking for a function $f$ such that for all $p$ and $q$... In other words, are you saying that $f$ does not depend on $p$ and $q$? – Rory Daulton Jun 15 '15 at 10:42
  • 2
    @RoryDaulton: OP said that $f(p)$ and $f(q)$ should be coprime for all such $p,q$. Though poorly stated, this means that it should do this for all coprime numbers. –  Jun 15 '15 at 10:45
  • Indeed, I think that if he meant $f_p(p)$ and $f_q(q)$ he would have written it, and it would be a very unusual question? – DanielV Jun 15 '15 at 10:47
  • 2
    @BolzWeir: I agree that seems to be what the OP means, but if so the question should have the introductory "If $p,q$ are co-prime integers, then" removed. I would do that edit myself if the OP himself made his meaning clear. I hate to edit other people's questions unless the meaning is made perfectly clear. – Rory Daulton Jun 15 '15 at 10:47
  • @RoryDaulton please edit question, I mean for all such coprimes, not restricted to only two numbers as in answer below. – hanugm Jun 15 '15 at 10:52
  • How about f(x) = x? – Esteban Crespi Jun 15 '15 at 13:47
  • 1
    @EstebanCrespi The question explicitly excludes $f(x) = x^n$ for any $n \geq 0$... – A.P. Jun 15 '15 at 14:00
  • I see no strong reason for closing this question. – GPerez Jun 15 '15 at 19:27

2 Answers2

1

How about $f(x)=(x-p)(x-q)+1$?

Anurag A
  • 41,067
  • 2
    I believe he means $\forall p, q$ – DanielV Jun 15 '15 at 10:18
  • In this case, $\mathrm{gcd}(f(p),f(q)) = \mathrm{gcd}((p-p)(p-q)+1,,(q-p)(q-q)+1) = \mathrm{gcd}(1,1) = 1$, so I believe it is true for all $p,q$. – molarmass Jun 15 '15 at 10:28
  • @molarmass Going from "$f(p)$ and $f(q)$ are coprime" to "$f(a)$ and $f(b)$ are coprime for every coprime pair $a,b$" is quite a big leap... – A.P. Jun 15 '15 at 10:50
  • $f(p)$ and $f(q)$ are coprime for all $p,q$, which implies $f(a)$ and $f(b)$ are coprime for every coprime pair $a,b$. – molarmass Jun 15 '15 at 10:53
  • 2
    @molarmass The definition of $f$ depends on $p$ and $q$. Thus you can't say "for all $p,q$" because those numbers are fixed as soon as you define $f$. – A.P. Jun 15 '15 at 11:05
1

Yeah, I realize I'm a bit late for this, but here goes:

We show that for any prime $p$, $P(p)$ is a (positive or negative) power of $p$. Assume for the sake of contradiction that some prime $q\neq p$ divides $P(p)$. Then

$$q|P(p)\implies q|P(p+q),$$

so $\gcd(P(p),P(p+q))\neq 1$, a contradiction. Now, as $-x^{d+1}<P(x)<x^{d+1}$ for all sufficiently large $x$ (if $d$ is the degree of $P$), we must have by the Pigeonhole principle that there exist some fixed $s\in\{-1,1\},k\in\mathbb{Z}_{\geq 0}$ so that $P(p)=sp^k$ for infinitely many primes $p$ (as the only possibilities for large enough $p$ are $\pm 1,\pm p,\cdots,\pm p^d$). But then

$$P(x)-sx^k$$

has infinitely many roots, and thus $P(x)$ is identically the polynomial $\pm x^k$.

Carl Schildkraut
  • 32,931