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Difference of Two Squares

Date: 09/04/2002 at 15:07:16
From: Steven
Subject: Difference of Two Squares

Hi,

I understand how to get the difference of two squares, but how would 
you get the equation when you have the answer?

Example:

   a^2 - b^2 = 45

Thanks,
Steven


Date: 09/05/2002 at 14:44:20
From: Doctor Greenie
Subject: Re: Difference of Two Squares

Hello, Steven -

I assume what you mean to ask is how you determine the possible values 
of a and b. I see two distinctly different ways.

With one approach, you use the equation as it stands and look for two 
perfect squares whose difference is 45.

With the other approach, you think of the equation in the form

  (a+b)(a-b) = 45

and try to find two numbers of the form (a+b) and (a-b) whose product 
is 45.

Using the first approach on your particular problem, we try successive 
values for b and see if the resulting value for a is an integer:

  b=1  ==>  a^2-1=45  ==>  a^2=46  no...
  b=2  ==>  a^2-4=45  ==>  a^2=49  yes; (a,b) = (7,2)
  b=3  ==>  a^2-9=45  ==>  a^2=54  no...
  b=4  ==>  a^2-16=45  ==>  a^2=61  no...
  b=5  ==>  a^2-25=45  ==>  a^2=70  no...
  b=6  ==>  a^2-36=45  ==>  a^2=81  yes; (a,b) = (9,6)
  b=7  ==>  a^2-49=45  ==>  a^2=94  no...
  b=8  ==>  a^2-64=45  ==>  a^2=109  no...
 ...

Clearly this way is going to be tedious, so let's look at the other 
approach.

  45 = (45)(1) = (23+22)(23-22) ==> (a,b) = (23,22)
  45 = (15)(3) = (9+6)(9-6) ==> (a,b) = (9,6)
  45 = (9)(5) = (7+2)(7-2) ==> (a,b) = (7,2)

These are the only three ways that 45 factors as the product of two 
integers, so these are the only three solutions for this example.

Of course, if you aren't just looking for integer answers, then there 
are many more (in fact an infinite number of) solutions. For example,

  45 = (10)(4.5) = (7.25+2.75)(7.25-2.75) ==> (a,b) = (7.25,2.75)

It might also be informative to look at another example using this 
second method, in which only some of the possible factorizations lead 
to integer solutions. If we consider, for example

  a^2-b^2 = 72

then we have, for the possible factorizations of 72 as the product of 
two integers:

  72 = (72)(1)
  72 = (36)(2)
  72 = (24)(3)
  72 = (18)(4)
  72 = (12)(6)
  72 = (9)(8)

The only factorizations that will lead to integer solutions for (a,b) 
are those in which either both factors are even or both are odd. For 
example, the factorization

  72 = (9)(8)

does not lead us to a solution in integers:

  72 = (9)(8) = (8.5+0.5)(8.5-0.5) ==> (a,b) = (8.5,0.5)

In practice, the first method might find you one or two solutions in 
a particular case, but the second method is generally much faster if 
you need to find all the solutions.

As an example where the first method is useful, consider

  a^2-b^2 = 3599

If you try the (normally better) second method on this one, you need 
to find the prime factorization of 3599, which for most people is 
going to be very time-consuming.  However, if you try the first 
method, you immediately see

  b=1  ==>  a^2-1=3599  ==>  a^2=3600  ==> (a,b) = (60,1)

This also shows you that the (prime) factorization of 3599 is

  3599 = (60+1)(60-1) = (61)(59)

I hope all this helps.  Please write back if you have any further 
questions about this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials
Middle School Factoring Expressions
Middle School Factoring Numbers

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