Difference between revisions of "Simon's Favorite Factoring Trick"

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=== Statement of the factorization ===
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Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user ComplexZeta, whose name is Simon. [http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=factoring&t=8215 This] appears to be the thread where '''Simon's favorite factoring trick''' was first introduced.
Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization. SFFT is: <math>{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)</math>.
 
  
=== Credit ===
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== Statement of the factorization ==
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The general statement of SFFT is: <math>{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)</math>.  More oftenly however SFFT is introduced as <math>xy + x + y + 1 = (x+1)(y+1)</math> or <math> xy - x - y +1 = (x-1)(y-1)</math>. 
  
This factorization was first popularized by AoPS user ComplexZeta, whose name is Simon.
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== Applications ==
 
 
=== Applications ===
 
 
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>{x}</math> and <math>{y}</math> are variables and <math>j,k</math> are known constants. Also it is typically necessary to add the <math>{j}{k}</math> term to both sides to perform the factorization.
 
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>{x}</math> and <math>{y}</math> are variables and <math>j,k</math> are known constants. Also it is typically necessary to add the <math>{j}{k}</math> term to both sides to perform the factorization.
  
=== Examples ===
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== Examples ==
 
([[AIME]] 1987/5) <math>m</math> and <math>n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
 
([[AIME]] 1987/5) <math>m</math> and <math>n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
  
'''Outline Solution:''' Rearrange to <math>m^2 + 3m^2n^2 -30n^2= 517</math>. The key step is changing the equation to <math>m^2 + 3m^2n^2 -30n^2-10= 507</math>, where the equation factors to <math>(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2</math>, from which the problem is trivial to solve by applying some simple number theory.
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'''Outline Solution:''' Rearrange to <math>m^2 + 3m^2n^2 -30n^2= 517</math>. The key step is changing the equation to <math>m^2 + 3m^2n^2 -30n^2-10= 507</math>, where the equation factors to <math>(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2</math>.  This makes things much simpler.  The rest of the problem is left as an exercise to the reader.

Revision as of 11:39, 20 June 2006

Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user ComplexZeta, whose name is Simon. This appears to be the thread where Simon's favorite factoring trick was first introduced.

Statement of the factorization

The general statement of SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$. More oftenly however SFFT is introduced as $xy + x + y + 1 = (x+1)(y+1)$ or $xy - x - y +1 = (x-1)(y-1)$.

Applications

This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually ${x}$ and ${y}$ are variables and $j,k$ are known constants. Also it is typically necessary to add the ${j}{k}$ term to both sides to perform the factorization.

Examples

(AIME 1987/5) $m$ and $n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.

Outline Solution: Rearrange to $m^2 + 3m^2n^2 -30n^2= 517$. The key step is changing the equation to $m^2 + 3m^2n^2 -30n^2-10= 507$, where the equation factors to $(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2$. This makes things much simpler. The rest of the problem is left as an exercise to the reader.

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