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Perfect Square Equation


Date: 02/22/2002 at 09:10:25
From: Barry Hynes
Subject: Prove perfect square equation

Could somebody explain the following equation for me, please?

  Prove that if n > 1, then nC2 + (n-1)C2 is a perfect square.

For n = 2 the answer is 1; for n = 3 the answer is 3 + 1 = 4; and for 
n = 4 the answer is 6 + 3 = 9.

How do you prove this? Through Sigma or by expanding nC2 into
n(n-1)!/2!*(n-2)!...

Barry


Date: 02/22/2002 at 11:23:53
From: Doctor Paul
Subject: Re: Prove perfect square equation

My first instinct is to say that using factorials is the way to go, 
although as I don't know if this will actually work. Let's see what 
happens:

C(n,2) + C(n-1,2) =

    n!          (n-1)!
---------  +  --------- = 
2! (n-2)!     2! (n-3)!


n*(n-1)/2 + (n-1)*(n-2)/2 = 

n*(n-1) + (n-1)*(n-2)
--------------------- = 
          2

(2*n^2 - 4*n + 2)/2 = n^2 - 2*n + 1 = (n-1)^2

So this seems to be the way to go. I'm sure there are other ways to 
prove this identity, but this one was the most obvious to me.

I hope this helps. Please write back if you'd like to talk about this 
more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   


Date: 02/24/2002 at 09:06:09
From: Barry Hynes
Subject: Prove perfect square equation

Hi,

Thanks for the help. The way you solved it was the way I was trying to 
do it, but I had the following and cannot see how you got to 

   n*(n-1)/2 + (n-1)*(n-2)/2 = 

What happened to the ! ? My math skills have evaporated over the past 
15 years.

MY WAY
   n!          (n-1)!
---------  +  --------- = 
2! (n-2)!     2! (n-3)!


 n(n-1)!    +  (n-1)(n-2)!
----------     -----------
(2!)(n-2)!     (2!)(n-3)!
---------------------------------------

YOUR WAY

    n!          (n-1)!
---------  +  --------- = 
2! (n-2)!     2! (n-3)!


n*(n-1)/2 + (n-1)*(n-2)/2 = 

Thanks,
Barry


Date: 02/24/2002 at 16:25:36
From: Doctor Paul
Subject: Re: Prove perfect square equation

                     5*4*3*2*1
Consider 5! / 3! =   --------- = 5*4
                       3*2*1

                        10*9*8*7*6*5*4*3*2*1
What about 10! / 8! ?   -------------------- = 10*9
                          8*7*6*5*4*3*2*1

We can generalize this:

               n * (n-1) * (n-2)!
n! / (n-2)! =  ------------------ =  n * (n-1)
                     (n-2)!

Does this help?  Please write back if you'd like to talk about this 
more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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