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

(Statement of the factorization)
(Applications)
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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.
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=== Examples ===
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([[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>.
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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>, from which the problem is trivial to solve by applying some simple number theory.

Revision as of 21:53, 17 June 2006

Statement of the factorization

Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization. SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$.

Credit

This factorization was first popularized by AoPS user ComplexZeta, whose name is Simon.

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 $(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2$, from which the problem is trivial to solve by applying some simple number theory.

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