Simon's Favorite Factoring Trick
Contents
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: Where
is the constant term,
is the product of the variables,
and
are the variables in linear terms.
Let's put it in general terms. We have an equation , where
,
, and
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:
Using the previous example,
is the same as:
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 and
are variables and
are known constants. Also, it is typically necessary to add the
term to both sides to perform the factorization.
Fun Practice Problems
Introductory
- Two different prime numbers between
and
are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
(Source)
Intermediate
Problem 1
- If
has a remainder of
when divided by
, and
has a remainder of
when divided by
, find the value of the remainder of when
is divided by
.
- icecreamrolls8
Solution
We have solution . Note that
can be factored into
using Simon's Favorite Factoring Trick. Now, look at n. Then, since the problem tells us that
has a remainder of
when divided by 5, we see that the
factor in the
expression has a remainder of
when divided by 5. Now, the
must have a remainder of
when divided by
as well (because then the main expression has a remainder of
when divided by
). Therefore, since 54 has a remainder of
when divided by
,
must have a remainder of
, our answer, so that the entire factor has a remainder of
when divided by
.
- icecreamrolls8
Hello Icecream rolls, your solution is likely flawed. The answer is E, as if you plug in 4 with all of your other statement, everything checks out. And your solution also has a ton of problems. Try to look into why or ask me to post my solution :) - Avatar of Wrath
Sorry to gang up on you, but is this really the best place to put this problem? In addition to being wrong, 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. Also, please replace '(first number) has a remainder of (second number) when divided (third number)' with I might also suggest moving this to a page on modulos. (or create one if it doesn't exist).
Problem 2
are integers such that
. Find
.
(Source)
Olympiad
- The integer
is positive. There are exactly
ordered pairs
of positive integers satisfying:
Prove that is a perfect square.
Source: (British Mathematical Olympiad Round 3, 2005)