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Ring Theoretical Method of Solving a Math Olympiad Problem

Math Olympiad problemCzech Republic Math Olympiad 2008 Problemquestion from russian math olympiadMath Olympiad Algebraic QuestionMath Olympiad Divisor ProblemMath Olympiad Perfect Square QuestionMath Olympiad Algebra QuestionDouble Integral Math Olympiad ProblemMath Olympiad Question- ProofMath Olympiad Problem regarding sum of digits.

1

2

$begingroup$

These paragraphs are from Steve Olson's book Count Down: Six Kids Vie for Glory at the World's Toughest Math Competition.

On page 170, the author said:

The sixth and last problem on the Forty-second Olympiad ---
by tradition the hardest of all --- looked deceptively straightforward to the competitors.

Let $a > b > c> d$ be positive integers and suppose that $ac +bd =(b + d + a -c)(b + d -a + c)$. Prove that $ab + cd$ is not
prime.

On page 174:

Gabriel's answer to problem six demonstrated his power as a
mathematician. "Gabe's solution was overkill," says Stankova,
"but he solved the problem the way a mathematician would solve it." In his solution he used a mathematical idea called a ring --- a set of mathematical objects, any two of which can be added or multiplied to yield another member of the set.

Edit
On page 210:

Gabriel's use of imaginary numbers in problem six was directly linked to the famous equation $e^pi i = -1$. The number omega ($omega$)
is defined as $omega = e^2pi i/3$. So $omega^2 = e^2pi i/3times e^2pi i/3=e^2pi i/3+2pi i/3=e^4pi i/3$ (because the exponents of $e$ can be added together when the two numbers are multiplied). By the same token, $omega^3 =e^6pi i/3=e^2pi i=e^pi itimes e^pi i= -1 times -1 = 1$.
Thus the set of numbers $1, -1, omega, -omega, omega^2$, and $-w^2$ are related in a particular way. If you multiply any two of them together, you get another member of the set. Gabriel used the powerful properties of this group to crack problem six.

The original problem: https://www.imo-official.org/problems.aspx (2001)

My Question: How did he solve the problem in a ring theoretical method?

ring-theory contest-math

share|cite|improve this question

edited Mar 28 at 17:02

bfhaha

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way that text is written is ... cringeworthy at best. Honestly, with problems like these, any methods that use ring properties can also be solved with the equivalent integer manipulations. I don't know what the author is getting at.
  $endgroup$
  – Don Thousand
  Mar 28 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe just working with $(a+c)(b+d)pmod ab+cd$? I didn't work it out.
  $endgroup$
  – rschwieb
  Mar 28 at 16:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also $(a-c)(b-d)$ might be important, because if $ab+cd$ is assumed to be prime, it is associate to the other product I mentioned in the quotient ring.
  $endgroup$
  – rschwieb
  Mar 28 at 16:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I forgot to quote a paragraph. I have made it. It seems to use group theory.
  $endgroup$
  – bfhaha
  Mar 28 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Af first glance, it seems to hint at employing the ring of Eisenstein integers. And a quick glance at an AoPS solution shows (unmotivated) integer calculations involving obvious Eisenstein integer norms, so it should be straightforward to reformulate that more naturally in Eisenstein arithmetic. Likely that is the "ring theoretic" solution implied in the book.
  $endgroup$
  – Bill Dubuque
  Mar 28 at 18:33
  
  

  
  
  
  

add a comment | 

1

2

$begingroup$

These paragraphs are from Steve Olson's book Count Down: Six Kids Vie for Glory at the World's Toughest Math Competition.

On page 170, the author said:

The sixth and last problem on the Forty-second Olympiad ---
by tradition the hardest of all --- looked deceptively straightforward to the competitors.

Let $a > b > c> d$ be positive integers and suppose that $ac +bd =(b + d + a -c)(b + d -a + c)$. Prove that $ab + cd$ is not
prime.

On page 174:

Gabriel's answer to problem six demonstrated his power as a
mathematician. "Gabe's solution was overkill," says Stankova,
"but he solved the problem the way a mathematician would solve it." In his solution he used a mathematical idea called a ring --- a set of mathematical objects, any two of which can be added or multiplied to yield another member of the set.

Edit
On page 210:

Gabriel's use of imaginary numbers in problem six was directly linked to the famous equation $e^pi i = -1$. The number omega ($omega$)
is defined as $omega = e^2pi i/3$. So $omega^2 = e^2pi i/3times e^2pi i/3=e^2pi i/3+2pi i/3=e^4pi i/3$ (because the exponents of $e$ can be added together when the two numbers are multiplied). By the same token, $omega^3 =e^6pi i/3=e^2pi i=e^pi itimes e^pi i= -1 times -1 = 1$.
Thus the set of numbers $1, -1, omega, -omega, omega^2$, and $-w^2$ are related in a particular way. If you multiply any two of them together, you get another member of the set. Gabriel used the powerful properties of this group to crack problem six.

The original problem: https://www.imo-official.org/problems.aspx (2001)

My Question: How did he solve the problem in a ring theoretical method?

ring-theory contest-math

share|cite|improve this question

edited Mar 28 at 17:02

bfhaha

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way that text is written is ... cringeworthy at best. Honestly, with problems like these, any methods that use ring properties can also be solved with the equivalent integer manipulations. I don't know what the author is getting at.
  $endgroup$
  – Don Thousand
  Mar 28 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe just working with $(a+c)(b+d)pmod ab+cd$? I didn't work it out.
  $endgroup$
  – rschwieb
  Mar 28 at 16:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also $(a-c)(b-d)$ might be important, because if $ab+cd$ is assumed to be prime, it is associate to the other product I mentioned in the quotient ring.
  $endgroup$
  – rschwieb
  Mar 28 at 16:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. I forgot to quote a paragraph. I have made it. It seems to use group theory.
  $endgroup$
  – bfhaha
  Mar 28 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Af first glance, it seems to hint at employing the ring of Eisenstein integers. And a quick glance at an AoPS solution shows (unmotivated) integer calculations involving obvious Eisenstein integer norms, so it should be straightforward to reformulate that more naturally in Eisenstein arithmetic. Likely that is the "ring theoretic" solution implied in the book.
  $endgroup$
  – Bill Dubuque
  Mar 28 at 18:33
  
  

  
  
  
  

add a comment | 

$begingroup$

These paragraphs are from Steve Olson's book Count Down: Six Kids Vie for Glory at the World's Toughest Math Competition.

On page 170, the author said:

The sixth and last problem on the Forty-second Olympiad ---
by tradition the hardest of all --- looked deceptively straightforward to the competitors.

Let $a > b > c> d$ be positive integers and suppose that $ac +bd =(b + d + a -c)(b + d -a + c)$. Prove that $ab + cd$ is not
prime.

On page 174:

Gabriel's answer to problem six demonstrated his power as a
mathematician. "Gabe's solution was overkill," says Stankova,
"but he solved the problem the way a mathematician would solve it." In his solution he used a mathematical idea called a ring --- a set of mathematical objects, any two of which can be added or multiplied to yield another member of the set.

Edit
On page 210:

Gabriel's use of imaginary numbers in problem six was directly linked to the famous equation $e^pi i = -1$. The number omega ($omega$)
is defined as $omega = e^2pi i/3$. So $omega^2 = e^2pi i/3times e^2pi i/3=e^2pi i/3+2pi i/3=e^4pi i/3$ (because the exponents of $e$ can be added together when the two numbers are multiplied). By the same token, $omega^3 =e^6pi i/3=e^2pi i=e^pi itimes e^pi i= -1 times -1 = 1$.
Thus the set of numbers $1, -1, omega, -omega, omega^2$, and $-w^2$ are related in a particular way. If you multiply any two of them together, you get another member of the set. Gabriel used the powerful properties of this group to crack problem six.

The original problem: https://www.imo-official.org/problems.aspx (2001)

My Question: How did he solve the problem in a ring theoretical method?

ring-theory contest-math

share|cite|improve this question

edited Mar 28 at 17:02

bfhaha

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

$endgroup$

share|cite|improve this question

edited Mar 28 at 17:02

bfhaha

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

share|cite|improve this question

edited Mar 28 at 17:02

bfhaha

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125

asked Mar 28 at 16:23

bfhahabfhaha

1,5751125