Rearranging a PolynomialDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/