Rearranging a Polynomial

Date: 02/23/2002 at 11:07:29
From: Amanda
Subject: Pythagorean Theorem

Hello.

How do you do this question?

   2n^2 + y^2 = 66^2

Thank you so much!
Mandy

Date: 02/23/2002 at 12:10:26
From: Doctor Schwa
Subject: Re: Pythagorean Theorem

Hi Mandy,

my first idea is to rearrange a bit:

   2n^2 = 66^2 - y^2,

and then

   2n^2 = (66-y)(66+y).

Then, since the left side is even, the right side is even too, so y 
must be even, call it 2r:

   2n^2 = (66-2r)(66+2r) = 4(33-r)(33+r).

Thus

    n^2 = 2(33-r)(33+r).

Now, since the right side is even, the left side must be even too, so 
n must be even, call it 2j:

   4j^2 = 2(33-r)(33+r)

   2j^2 = (33-r)(33+r).

Now, since the left side is even, the right side is even too, so r 
must be odd, call it 2s+1:

   2j^2 = (33-2s-1)(33+2s+1)

   2j^2 = (32-2s)(34+2s) = 4(16-s)(18+s), so

    j^2 = 2(16-s)(18+s)

By now the problem is getting small enough that you could try all the 
possible values of s (0 through 16, with 0 and 16 giving you the silly 
solution n = 0, y = 66 to your original problem), or you could keep 
applying the same idea (now j has to be even, so call j = 2k, and 
substitute yet again...)

Enjoy,

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