2006 USAMO Problems/Problem 3: Difference between revisions

Revision as of 19:37, 1 September 2006

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

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 == Problem ==
-For integral <math>m</math>, let <math>p(m)</math> be the greatest prime divisor of <math>m</math>. By convention, we set <math>p(\pm 1)=1</math> and <math>p(0)=\infty</math>. Find all polynomial <math>f</math> with integer coefficients such that the sequence
-<math>(p(f(n^2))-2n)_{n\ge 0}</math>
+For integral <math>\displaystyle m </math>, let <math> \displaystyle p(m) </math> be the greatest prime divisor of <math> \displaystyle m </math>. By convention, we set <math> p(\pm 1)=1</math> and <math>p(0)=\infty</math>. Find all polynomials <math>\displaystyle f </math> with integer coefficients such that the sequence <math> \{ p(f(n^2))-2n) \} _{n\ge 0} </math> is bounded above. (In particular, this requires <math>f(n^2)\neq 0</math> for <math>n\ge 0</math>.)
-is bounded above. (In particular, this requires <math>f(n^2)\neq 0</math> for <math>n\ge 0</math>)
 == Solution ==
+{{solution}}
 == See Also ==
-*[[2006 USAMO Problems]]
+* [[2006 USAMO Problems]]
+* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=490625#p490625 Discussion on AoPS/MathLinks]
+[[Category:Olympiad Number Theory Problems]]

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2006 USAMO Problems/Problem 3: Difference between revisions

Revision as of 19:37, 1 September 2006

Problem

Solution

See Also