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我们知道,3~2+4~2=5~2。20~2+21~2=29~2,119~2+120~2=169~2等等,有许多这样的前两数为连续整数的勾股数组,它们如何求出来?这个问题实际上就是求不定方程x~2+(x+1)~2=z~2的所有正整数解。定理:x~2+(x+1)~2=z~2的所有正整数解是: 证明:∵x~2+(x+1)~2=z~2 ∴2x~2+2x+1=z~2 ∴(2x+1)~2+1=2z~2 即 (2x+1)~2-2z~2=-1 ∴ (2x+1+2~(1/2)z)(2x+1-(2z)~(1/2)) =-1因为有(1+2~(1/2))(1-2~(1/2))=-1
We know that 3~2+4~2=5~2. 20~2+21~2=29~2,119~2+120~2=169~2 and so on. There are many such first two numbers as continuous integers. The Pythagorean arrays, how are they solved? The problem is actually to find all the positive integer solutions of the indefinite equation x~2+(x+1)~2=z~2. Theorem: All positive integer solutions of x~2+(x+1)~2=z~2 are: Proof: ∵x~2+(x+1)~2=z~2 ∴2x~2+2x+1 =z~2 ∴(2x+1)~2+1=2z~2 (2x+1)~2-2z~2=-1 ∴ (2x+1+2~(1/2)z)(2x +1-(2z)~(1/2)) =-1 because (1+2~(1/2))(1-2~(1/2))=-1