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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.  
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The answer is the set of all integers that are at least <math>3</math>.
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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:
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Pick maximal integers j_1 and j_2 such that ((p_1)^(j_1))((p_2)^(j_2)) divides i.  
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For composite <math>n</math> where there are two primes <math>p_1</math> and <math>p_2</math> such that <math>\frac{n}{2}<p_1<p_2<n</math>, here's your construction:
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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.)
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Pick maximal integers <math>j_1</math> and <math>j_2</math> such that <math>p_1^{j_1}p_2^{j_2}</math> divides <math>i</math>.  
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Then a_i=(s^(j_1))(t^(j_2)).
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Pick a minimal positive integer s such that <math>\frac{n(n+1)}{2}+(s-1)p_1 \equiv 0</math> (mod <math>p_2</math>). (You know it exists since <math>p_1</math> and <math>p_2</math> are relatively prime.)
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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.)
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Then <math>a_i=(s^{j_1})(t^{j_2})</math>.
  
 
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.
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<math>a_i=(-1)^{j_1+j_2}</math>, where<math>2^{j_1}3^{j_2} </math>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.
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<math>a_i=2^{j_1}(-5)^{j_2}</math>, where <math>3^{j_1}5^{j_2}</math>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.
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<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)

Latest revision as of 15:22, 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 $\frac{n}{2}<p_1<p_2<n$, 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 $\frac{n(n+1)}{2}+(s-1)p_1 \equiv 0$ (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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