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### 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/
```
Associated Topics:
High School Polynomials

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