2006 USAMO Problems/Problem 3

Problem

(Titu Andreescu, Gabriel Dospinescu) For integral $m$ , let $p(m)$ be the greatest prime divisor of $m$ . By convention, we set $p(\pm 1)=1$ and $p(0)=\infty$ . Find all polynomials $f$ with integer coefficients such that the sequence $\{ p(f(n^2))-2n) \}_{n\ge 0}$ is bounded above. (In particular, this requires $f(n^2)\neq 0$ for $n\ge 0$ .)

Solutions

Solution 1

Let $f(x)$ be a non-constant polynomial in $x$ of degree $d$ with integer coefficients, suppose further that no prime divides all the coefficients of $f$ (otherwise consider the polynomial obtained by dividing $f$ by the gcd of its coefficients). We further normalize $f$ by multiplying by $-1$ , if necessary, to ensure that the leading coefficient (of $x^d$ ) is positive.

Let $g(n) = f(n^2)$ , then $g(n)$ is a polynomial of degree $2$ or more and $g(n) = g(-n)$ . Let $g_1, \ldots, g_k$ be the factorization of $g$ into irreducible factors with positive leading coefficients. Such a factorization is unique. Let $d(g_i)$ denote the degree of $g_i$ . Since $g(-n) = g(n)$ the factors $g_i$ are either even functions of $n$ or come in pairs $(g_i, h_i)$ with $g_i(-n) = (-1)^{d(g_i)} h_i(n)$ .

Let $P(0) = \infty$ , $P(\pm 1) = 1$ . For any other integer $m$ let $P(m)$ be the largest prime factor of $m$ .

Suppose that for some finite constant $C$ and all $n \ge 0$ we have $P(g(n)) - 2n < C$ . Since the polynomials $g_i$ divide $g$ , the same must be true for each of the irreducible polynomials $g_i$ .

A theorem of T. Nagell implies that if $d(g_i) \ge 2$ the ratio $P(g_i(n))/n$ is unbounded for large values of $n$ . Since in our case the $P(g_i(n))/n$ is asymptotically bounded above by $2$ for large $n$ , we conclude that all the irreducible factors $g_i$ are linear. Since linear polynomials are not even functions of $n$ , they must occur in pairs $g_i(n) = a_in + b_i$ , $h_i(n) = a_in - b_i$ . Without loss of generality, $b_i \ge 0$ . Since the coefficients of $f$ are relatively prime, so are $a_i$ and $b_i$ , and since $P(0) = \infty$ , neither polynomial can have any non-negative integer roots, so $a_i > 1$ and thus $b_i > 0$ .

On the other hand, by Dirichlet's theorem, $a_i \le 2$ , since otherwise the sequence $a_in + b_i$ would yield infinitely many prime values with $P(g_i(n)) = a_in + b_i \ge 3n.$ So $a_i = 2$ and therefore $b_i$ is a positive odd integer. Setting $b_i = 2c_i + 1$ , clearly $P(g_i(n)) - 2n < 2c_i + 2$ . Since this holds for each factor $g_i$ , it is true for the product $g$ of all the factors with the bound determined by the factor with the largest value of $c_i$ .

Therefore, for suitable non-negative integers $c_i$ , $g(n)$ is a product of polynomials of the form $4n^2 - (2c_i + 1)^2$ . Now, since $g(n) = f(n^2)$ , we conclude that $f(n)$ is a product of linear factors of the form $4n - (2c_i + 1)^2$ .

Since we restricted ourselves to non-constant polynomials with relatively prime coefficients, we can now relax this condition and admit a possibly empty list of linear factors as well as an arbitrary non-zero integer multiple $M$ . Thus for a suitable non-zero integer $M$ and $k \ge 0$ non-negative integers $c_i$ , we have: $f(n) = M \cdot \prod_{i=1}^k (4n - (2c_i + 1)^2)$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2006 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search

2006 USAMO Problems/Problem 3

Contents

Problem

Solutions

Solution 1

See also