10.14

Altheman

Find all quadratic polynomials $f\in \mathbb{Z}[X]$ with
the property that for any relatively prime integers $m,n$ , the
numbers $f(m),f(n)$ are also relatively prime.

**Posted:** Tue Jul 21, 2009 9:36 pm

_________________
-Alex
Altheman's Problem Column

azjps

Suppose $f(a) = b$ with $|\text{gcd}\,(a,b)| = 1$ , then $b|f(a + b) - f(a) \Longrightarrow 1 = |\text{gcd}\,(f(a + b),f(a))| \ge |b|$ , so $b \in \{ - 1,1\}$ (certainly $b \neq 0$ ). It follows that $f( - 1),f(1) = \pm 1$ and $f(p) = \pm p^k$ for any prime $p$ . Then $f(p) = c_2p^2 + c_1p + f(0) = \pm p^k$ implies that $f(0) \equiv \{ - 1,0,1\} \pmod{p}$ for all $p$ . Clearly then $f(0) \in \{ - 1,0,1\}$ . We quickly get the possible solutions $f(x) = \pm \{x^2, 2x^2 - 1, x^2 - x - 1, x^2 + x - 1, x, 1\}$ of which only $f(x) = x^2$ satisfies the conditions.

**Posted:** Wed Jul 29, 2009 7:18 am

harazi

The same argument actually gives all polynomials, not only quadratic ones: the main point is that for $n$ large enough we cannot have $(n,f(n))=1$ , otherwise by picking $k$ large such that $(n,n+kf(n))=1$ we get a contradiction. But $(n,f(n))\ne 1$ for all large $n$ implies $f(0)=0$ and now repeat the argument with $f(X)/X$ .

**Posted:** Wed Jul 29, 2009 5:41 pm

harazi

Here is a (very difficult for me) problem: find all $f: \{1,2,...\}\to \{1,2,...\}$ such that $m-n$ divides $f(m)-f(n)$ for all $m,n$ and such that for all $gcd(m,n)=1$ we also have $gcd(f(m),f(n))=1$ . I thought I solved it quite a long time, but then realized I was wrong. Someone has the courage to trivalize it? Mr. Green

**Posted:** Wed Nov 04, 2009 8:50 am

fedja

**Posted:** Wed Nov 04, 2009 5:01 pm