Polynomial divisibility

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#1 h Mar 29, 2018, 9:49 PM • 4 Y

Y by Lamp909, TwoTimes3TimesSeven, Adventure10, Mango247

Find all polynomials $f\in Z[X],$ such that for each odd prime $p$ $f(p)|(p-3)!+\frac{p+1}{2}.$

This post has been edited 1 time. Last edited by k.vasilev, Mar 29, 2018, 9:52 PM

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#2 h Mar 29, 2018, 11:55 PM • 3 Y

Y by k.vasilev, Adventure10, Mango247

Answer : $P(x)= \pm x, 1$ . If $P$ is constant then it is not hard to see $P=1$ .Assume $P(0) \neq 0$ .Let $p$ be a prime that does not divide $P(0)$ and $p/P(n)$ for some n .(Such exists by Shur's Lemma). Now we have $(n,p)=1$ and by Dirichlet's theorem there are infinitely many primes of the form $q=kp+n$ . Lets take one such $q$ that is sufficiently big. We have that $p/f(q)/(q-3)!+\frac{q+1}{2}$ . This means that $p/q+1$ . Whence $p/n+1$ . So $p / P(-1)$ for inf many $p$ ,so $P(-1)=0$ . So $p+1/P(p)/(p-3)!+\frac{p+1}{2}$ for all odd p. p=3 gives $4/3$ .Contradiction. So $P(0)=0$ and $p/P(p)$ for all odd primes $p$ . Now write $P(x)=xQ(x)$ and apply the same for $Q$ . This gives $P(x)=cx^{n}$ . Plugging $p=3$ gives $c=\pm 1$ and $n=1$ . Now what ramains is to check. Note that $(p-1)! \equiv -1(p)$ gives $(p-3)! \equiv \frac{-1}{2}(p)$ , which yields the result.

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#3 h Mar 30, 2018, 1:28 PM • 4 Y

Y by k.vasilev, Boris2005, Adventure10, Mango247

We know that there are infinitely many prime divisors of $f(x)$ and only finitely many primes that divide $f(0)$ , thus we will look at the infinitely many primes that divide $f(x)$ for some $x$ and don't divide $f(0)$ . Let's denote it by $q$ . So $q/f(n)$ for some $n$ . Now let's look at the arithm. seq. $qk+n$ and let's assume $(q;n)=1$ . Now, by Dirichlet we know there are infinitely many primes in it. Let's take a very very large one( $p-3>q$ ). $q/f(n) \Rightarrow q/f(p)/(p-3)!+\frac{p+1}{2} \Rightarrow q/p+1 \Rightarrow q/p+1=qk+n \Rightarrow q/n+1 \Rightarrow q/f(-1)$ But we know there are infinitely many such primes $q$ and only finitely many divisors of $f(-1)$ , thus we reach a contradiction with the assumption that $(q;n)=1. \Rightarrow q/n$ for every $q/f(n) \Rightarrow q/f(0)$ for infinitely many primes $q \Rightarrow f(0)=0$ . But we know that if $f(x)=xf_1(x)$ satisfies the condition, then $f_1(x)$ also does. Thus we can do this until we reduce it to a constant. $\Rightarrow f(x)=cx^n$ . Checking for $x=3 \Rightarrow c \cdot 3^n/3 \Rightarrow n=1$ and $c=1$ or $c=-1$ . So $f(x)=x$ or $f(x)=-x$ .

This post has been edited 2 times. Last edited by TwoTimes3TimesSeven, Mar 30, 2018, 1:31 PM

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