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Difference between revisions of "2006 USAMO Problems/Problem 3"

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non-negative integers <math>c_i</math>, we have:
 
non-negative integers <math>c_i</math>, we have:
 
<cmath>f(n) = M \cdot \prod_{i=1}^k (4n - (2c_i + 1)^2)</cmath>
 
<cmath>f(n) = M \cdot \prod_{i=1}^k (4n - (2c_i + 1)^2)</cmath>
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Note: a more elementary solution can be found at [http://www.williams.edu/go/math/sjmiller/public_html/greenchicken/otherexams/olympiad0607solns.pdf USA and International Mathematical Olympiads 2006-2007 by Zuming Feng and Yufei Zhao]
  
 
== See Also ==
 
== See Also ==

Revision as of 10:18, 26 March 2010

Problem

For integral $\displaystyle m$, let $\displaystyle p(m)$ be the greatest prime divisor of $\displaystyle m$. By convention, we set $p(\pm 1)=1$ and $p(0)=\infty$. Find all polynomials $\displaystyle f$ 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 $n\ge 0$.)

Solution

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 polynomials 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 $k \ge 0$ irreducible factors as well as an arbitrary non-zero integer multiple $M$. Thus for a suitable non-zero integer $M$ and a possibly empty set of $k$ non-negative integers $c_i$, we have: \[f(n) = M \cdot \prod_{i=1}^k (4n - (2c_i + 1)^2)\]

Note: a more elementary solution can be found at USA and International Mathematical Olympiads 2006-2007 by Zuming Feng and Yufei Zhao

See Also

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