Difference between revisions of "Talk:2012 USAMO Problems/Problem 3"

(Extension of the proof.)
(No difference)

Revision as of 09:00, 25 April 2012

The answer is the set of all integers that are at least 3. For composite n where there are two primes p_1 and p_2 such that n/2<p_1<p_2<n, here's your construction: Pick maximal integers j_1 and j_2 such that ((p_1)^(j_1))((p_2)^(j_2)) divides i. Pick a minimal positive integer s such that (n(n+1)/2)+(s-1)(p_1) is 0 mod p_2. (You know it exists since p_1 and p_2 are relatively prime.) Pick an integer t such that (n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0. (It exists because of how we defined s. It also must be negative.) Then a_i=(s^(j_1))(t^(j_2)).

For n=4: a_i=(-1)^(j_1+j_2), where (2^j_1)(3^j_2) divides i. For n=6: a_i=(2^j_1)(-5)^j_2, where (3^j_1)(5^j_2) divides i. For n=10: a_i=(2^j_1)(-9)^j_2, where (5^j_1)(7^j_2) divides i.

[I don't know LaTeX, so someone else can input it.] --Mage24365 09:00, 25 April 2012 (EDT)