# Simon's Favorite Factoring Trick

## The General Statement For Nerds

Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine(Positive) 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. A 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 integral constants. According to Simon's Favourite 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

## 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)

## Problem 1

• If $kn+54k+2n+106$ 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 $5$.

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 }$

## Solution

We have solution $3$. Note that $kn+53k+2n+106$ 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) NERD

• 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.