困難奧數一條，只求解釋！
回答 (2)
2010-07-26 4:06 pm
✔ 最佳答案
已知1889=33^2+2*20^2，且存在正整數x、y，使得1889^2=x^2+2y^2。求x、ySol
1889=33^2+2*20^2
=1089+2*400
=1089+800
=289+1600
1889^2=289^2+2*289*1600+1600^2
=289^2+1600*(2*289+1600)
=289^2+1600*2178
=289^2+2*1600*1089
=289^2+2*40^2*33^2
=289^2+2*1320^2
本題上述為唯一解答，理論上只可以說是一組解
2010-07-26 6:08 pm
The answer does not show how to proceed with this step:
=1089+800
=289+1600
denote a = 33^2, b = 20^2, we see
(a + 2b)^2 = a^2 + 4ab + 4b^2
= a^2 - 4ab + 4b^2 + 8ab
= (a - 2b)^2 + 2(4ab)
Remember a and b are squares, 4ab is also a square and the result follows.
=1089+800
=289+1600
denote a = 33^2, b = 20^2, we see
(a + 2b)^2 = a^2 + 4ab + 4b^2
= a^2 - 4ab + 4b^2 + 8ab
= (a - 2b)^2 + 2(4ab)
Remember a and b are squares, 4ab is also a square and the result follows.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100725000051KK01568