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

Revision as of 14:31, 8 August 2022

The General Statement

Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. An extortive example would be: $xy+66x-88y=23333$ where $23333$ is the constant term, $xy$ is the product of the variables, $66x$ and $-88y$ are the variables in linear terms.

Let's put it in general terms. We have an equation $xy+jx+ky=a$ , where $j$ , $k$ , and $a$ are integer constants, and the coefficient of xy must be 1(If it is not 1, then divide the coefficient off of the equation.). According to Simon's Favorite Factoring Trick, this equation can be transformed into: $(x+k)(y+j)=a+jk$ Using the previous example, $xy+66x-88y=23333$ is the same as: $(x-88)(y+66)=(23333)+(-88)(66)$

If this is confusing or you would like to know the thought process behind SFFT, see this eight-minute video by Richard Rusczyk from AoPS: https://www.youtube.com/watch?v=0nN3H7w2LnI. For the thought process, start from https://youtu.be/0nN3H7w2LnI?t=366.

Here's another way to look at it.

Consider the equation $xy+5x+6y=30$ .Let's start to factor the first group out: $x(y+5)+6y=30$ .

How do we group the last term so we can factor by grouping? Notice that we can add $30$ to both sides. This yields $x(y+5)+6(y+5)=60$ . Now, we can factor as $(x+6)(y+5)=60$ .

This is important because this keeps showing up in number theory problems. Let's look at this problem below:

Determine all possible ordered pairs of positive integers $(x,y)$ that are solutions to the equation $\frac{4}{x}+\frac{5}{y}=1$ . (2021 CEMC Galois #4b)

Let's remove the denominators: $4y+5x=xy$ . Then $xy-5x-4y=0$ . Take out the $x$ : $x(y-5)-4(y-5)=0+20$ (notice how I artificially grouped up the $y$ terms by adding $4*5=20$ ).

Now, $(x-4)(y-5)=20$ (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem.

Look at all factor pairs of 20: $1*20, 2*10, 4*5, 5*4, 10*2, 20*1$ . The first factor is for $x$ , the second is for $y$ . Solving for each of the equations, we have the solutions as $\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}$ .

For more info on the solution: https://www.cemc.uwaterloo.ca/contests/past_contests/2021/2021GaloisSolution.pdf

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 $jk$ term to both sides to perform the factorization.

Fun Practice Problems

Introductory

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$\mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$

(Source)

Intermediate

Problem 1

If $kn+54k+2n+108$ has a remainder of $4$ when divided by $5$ , and $k$ has a remainder of $1$ when divided by $5$ , find the value of the remainder of when $n$ is divided by $5$ .

$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 0 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 }$

- icecreamrolls8

Solution

We have solution $3$ . Note that $kn+54k+2n+108$ can be factored into $(k+2)(n+54)$ using Simon's Favorite Factoring Trick. Now, look at n. Then, since the problem tells us that $k$ has a remainder of $1$ when divided by 5, we see that the $(k+2)$ factor in the $(k+2)(n+54)$ expression has a remainder of $3$ when divided by 5. Now, the $(n+54)$ must have a remainder of $3$ when divided by $5$ as well (because then the main expression has a remainder of $4$ when divided by $5$ ). Therefore, since 54 has a remainder of $4$ when divided by $5$ , $n$ must have a remainder of $3$ , our answer, so that the entire factor has a remainder of $3$ when divided by $5$ .

- icecreamrolls8

Problem 2

$m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$ . Find $3m^2n^2$ .

(Source)

Olympiad

The integer $N$ is positive. There are exactly $2005$ ordered pairs $(x, y)$ of positive integers satisfying:

$\frac 1x +\frac 1y = \frac 1N$

Prove that $N$ is a perfect square.

Source: British Mathematical Olympiad Round 2 #1 https://bmos.ukmt.org.uk/home/bmo2-2005.pdf

See More

@@ Line 53: / Line 53: @@
 - icecreamrolls8
-Is this really the best place to put this problem? It makes a large jump, simply 'noting' that the equation can be factored a certain way, instead of explaining how to factor it. In my experience, this is the hardest part of learning Simon's Favorite Factoring Trick, and you skip through it. I might also suggest moving this to a page on modulos. (or create one if it doesn't exist).
 ==Problem 2==
@@ Line 68: / Line 67: @@
 Prove that <math>N</math> is a perfect square.
-Source: (British Mathematical Olympiad Round 3, 2005)
+Source: British Mathematical Olympiad Round 2 #1 https://bmos.ukmt.org.uk/home/bmo2-2005.pdf
 == See More==

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