Difference between revisions of "Talk:2012 USAMO Problems/Problem 3"
(Extension of the proof.) |
m |
||
Line 1: | Line 1: | ||
The answer is the set of all integers that are at least 3. | 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: | 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 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 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.) | 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)). | Then a_i=(s^(j_1))(t^(j_2)). | ||
For n=4: | For n=4: | ||
+ | |||
a_i=(-1)^(j_1+j_2), where (2^j_1)(3^j_2) divides i. | a_i=(-1)^(j_1+j_2), where (2^j_1)(3^j_2) divides i. | ||
+ | |||
For n=6: | For n=6: | ||
+ | |||
a_i=(2^j_1)(-5)^j_2, where (3^j_1)(5^j_2) divides i. | a_i=(2^j_1)(-5)^j_2, where (3^j_1)(5^j_2) divides i. | ||
+ | |||
For n=10: | For n=10: | ||
+ | |||
a_i=(2^j_1)(-9)^j_2, where (5^j_1)(7^j_2) divides i. | 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.] | [I don't know LaTeX, so someone else can input it.] | ||
+ | |||
--[[User:Mage24365|Mage24365]] 09:00, 25 April 2012 (EDT) | --[[User:Mage24365|Mage24365]] 09:00, 25 April 2012 (EDT) |
Revision as of 08:00, 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)