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Difference between revisions of "Talk:2012 USAMO Problems/Problem 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<math>((p_1)^(j_1))((p_2)^(j_2))</math> divides i.  
+
Pick maximal integers j_1 and j_2 such that<math>((p_1)^{j_1})((p_2)^{j_2})</math> divides i.  
  
−
Pick a minimal positive integer s such that <math> (n(n+1)/2)+(s-1)(p_1)</math> 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 <math>\frac{n(n+1)}{2}+(s-1)p_1</math> is <math>0</math> mod <math>p_2</math>. (You know it exists since p_1 and p_2 are relatively prime.)
  
−
Pick an integer t such that<math> (n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0</math>. (It exists because of how we defined s. It also must be negative.)
+
Pick an integer t such that<math> \frac{n(n+1)}{2}+(s-1)p_1+(t-1)p_2=0</math>. (It exists because of how we defined s. It also must be negative.)
  
−
Then <math>a_i=(s^(j_1))(t^(j_2))</math>.
+
Then <math>a_i=(s^{j_1})(t^{j_2})</math>.
  
 
For n=4:
 
For n=4:
  
−
<math>a_i=(-1)^(j_1+j_2)</math>, where<math> (2^j_1)(3^j_2) </math>divides i.
+
<math>a_i=(-1)^{j_1+j_2}</math>, where<math>2^{j_1}3^{j_2} </math>divides i.
  
 
For n=6:
 
For n=6:
  
−
<math>a_i=(2^j_1)(-5)^j_2</math>, where <math>(3^j_1)(5^j_2)</math>divides i.
+
<math>a_i=2^{j_1}(-5)^{j_2}</math>, where <math>3^{j_1}5^{j_2}</math>divides i.
  
 
For n=10:
 
For n=10:
  
−
<math>a_i=(2^j_1)(-9)^j_2</math>, where <math>(5^j_1)(7^j_2)</math>divides i.
+
<math>a_i=2^{j_1}(-9)^{j_2}</math>, where <math>5^{j_1}7^{j_2}</math>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 15:18, 3 May 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 $\frac{n(n+1)}{2}+(s-1)p_1$ is $0$ mod $p_2$. (You know it exists since p_1 and p_2 are relatively prime.)

Pick an integer t such that$\frac{n(n+1)}{2}+(s-1)p_1+(t-1)p_2=0$. (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)

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