求解方程组d^3-a^2*d^2-5*d-2*a=0,a*d^3+5*a*d-2*d+2*a^2=0

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d^3-a^2*d^2-5*d-2*a=0————（1）
a*d^3+5*a*d-2*d+2*a^2=0——————（2）

angel_phoenix99

楼主确认a^2-b^2=2ab没有非零整数解等价于费马定理?

a^2-b^2=2ab没有非零整数解的证明过程

若该方程有非零整数解(a，b) ，则必然存在互质基础解(m  n)=1
(a，b)=t>1
a=tm
b=tn
(m ，n)=1且m^2-n^2=2mn


下面来证明互质的(m，n)不存在
由于(m，n)=1，(m-2n，n)=1

m(m-2n)=n^2不可能成立

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b有唯一解
b= (((c*sqrt(a*x-d)-2*a^2*x^2+3*a*d*x-d^2)/(a*x-d)))
C有唯一解
c = (-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d))
d有唯一解d=(a*x-x^2)
a有唯一解a= ((x^2+d)/x)
(x^3+a*x^2+((((-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d))*sqrt(a*x-d)-2*a^2*x^2+3*a*d*x-d^2)/(a*x-d)))*x+(-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d)))*(x^2-a*x+d)   =


(x^3+a*x^2+b*x+c)*(x^2-a*x+d)   =
=x^5+(d+b-a^2)*x^3+(a*d+c-a*b)*x^2+(b*d-a*c)*x+c*d
=x^5+(d+b-a^2)*x^3+(a*d+(-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d))-a*b)*x^2+(b*d-a*(-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d)))*x+(-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d))*d
= b*x^3-a*b*x^2+b*d*x
结论：若b=0，则五次方程可解。
表达式 (x^3+a*x^2+c)*(x^2-a*x+d) = x^5+(d-a^2)*x^3+(a*d+c)*x^2-a*c*x+c*d
x^5+(d-a^2)*x^3+(a*d+c)*x^2-a*c*x+c*d=
令c = (2*a^2*x^2-3*a*d*x+d^2)/sqrt(a*x-d)
= (-(a*x^4+((-d)-a^2)*x^3+2*a*d*x^2-d^2*x)/(x*sqrt(a*x-d)+a*x-d))
令b= (((c*sqrt(a*x-d)-2*a^2*x^2+3*a*d*x-d^2)/(a*x-d)))=0

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若a^2-b^2=2ab没有整数解，则“费马大定理”得证。

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wangyangke

:):):)

wangyangke

:):):)

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wangyangke 发表于 2018-3-10 09:37

简单不？能用半页纸么？