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

(Extension of the proof.)
 
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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.  
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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.)
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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.)
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Then a_i=(s^(j_1))(t^(j_2)).
 
Then a_i=(s^(j_1))(t^(j_2)).
  
 
For n=4:
 
For n=4:
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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.
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For n=6:
 
For n=6:
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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.
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For n=10:
 
For n=10:
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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.]
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--[[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)