Difference between revisions of "2014 USAMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | + | The following solution is due to Gabriel Dospinescu and v_Enhance (also known as Evan Chen). | |
− | + | ||
+ | Let <math>N = n+1</math> and assume <math>N</math> is (very) large. We construct an <math>N \times N</math> with cells <math>(i,j)</math> where <math>0 \le i, j \le n</math> and in each cell place a prime <math>p</math> dividing <math>\gcd (a+i, b+j)</math>. | ||
+ | |||
+ | The central claim is at least <math>50\%</math> of the primes in this table exceed <math>0.001n^2</math>. We count the maximum number of squares they could occupy: | ||
+ | <cmath> | ||
+ | \sum_p \left\lceil \frac{N}{p} \right\rceil^2 | ||
+ | \le \sum_p \left( \frac Np + 1 \right)^2 \\ | ||
+ | = N^2 \sum_p \frac{1}{p^2} + 2N \sum_p \frac1p + \sum_p 1. | ||
+ | </cmath> Here the summation runs over primes <math>p \le 0.001n^2</math>. | ||
+ | |||
+ | Let <math>r = \pi(0.001n^2)</math> denote the number of such primes. Now we apply the three bounds: <cmath> \sum_p \frac{1}{p^2} < \frac 12 </cmath> which follows by adding all the primes directly with some computation, <cmath> \sum_p \frac 1p < \sum_{k=1}^r \frac 1k = O(\ln r) < o(N) </cmath> using the harmonic series bound, and <cmath> \sum_p 1 < r \sim O \left( \frac{N^2}{\ln N} \right) < o(N^2) </cmath> via Prime Number Theorem. Hence the sum in question is certainly less than <math>\tfrac 12 N^2</math> for <math>N</math> large enough, establishing the central claim. | ||
+ | |||
+ | Hence some column <math>a+i</math> has at least one half of its primes greater than <math>0.001n^2</math>. Because this is greater than <math>n</math> for large <math>n</math>, these primes must all be distinct, so <math>a+i</math> exceeds their product, which is larger than <cmath> \left( 0.001n^2 \right)^{N/2} > c^n \cdot n^n </cmath> where <math>c</math> is some constant. |
Latest revision as of 10:55, 25 June 2020
Problem
Prove that there is a constant with the following property: If are positive integers such that for all , then
Solution
The following solution is due to Gabriel Dospinescu and v_Enhance (also known as Evan Chen).
Let and assume is (very) large. We construct an with cells where and in each cell place a prime dividing .
The central claim is at least of the primes in this table exceed . We count the maximum number of squares they could occupy: Here the summation runs over primes .
Let denote the number of such primes. Now we apply the three bounds: which follows by adding all the primes directly with some computation, using the harmonic series bound, and via Prime Number Theorem. Hence the sum in question is certainly less than for large enough, establishing the central claim.
Hence some column has at least one half of its primes greater than . Because this is greater than for large , these primes must all be distinct, so exceeds their product, which is larger than where is some constant.