已知a+b=c+d,a³+b³=c³+d³,求证a2003次方+b2003次方=c2003次方+d2003次方

Question

已知a+b=c+d,a³+b³=c³+d³,求证a2003次方+b2003次方=c2003次方+d2003次方

Answer 1

归纳法,已知a+b=c+d,
(a+b)(a+b)=(c+d)(c+d) => a^2+2ab+b^2=c^2+2cd+d^2---(1)
a^3+b^3=c^3+d^3 => (a+b)(a^2-ab+b^2)=(c+d)(c^2-cd+d^2)---(2)
(1)(2)联立=> ab=cd => a^2+b^2 = c^2+d^2
(a+b)(a^3+b^3)=(c+d)(c^3+d^3) => a^4+ab^3+a^3b+b^4=c^4+cd^3+c^3d+d^4---(3)
(a^2+b^2)(a^2+b^2) = (c^2+d^2)(c^2+d^2) => ...=> a^4+2a^2b^2+b^4=c^4+2c^2d^2+d^4---(4)
(3)(4)联立=> ab^3+a^3b = cd^3+c^3d => a^4=d^4
假设自然数n,m,i,满足n>=m>=i,任意不大于n次的齐次二项式a^(m-i)b^i+a^ib^(m-i)=c^(m-i)d^i+c^id^(m-i)都成立；
当m=n,i=0时,有(a+b)(a^n+b^n)=(c+d)(c^n+d^n) => a^(n+1)+ab^n+a^nb+b^(n+1)=c^(n+1)+cd^n+c^nd+d^(n+1)---(5)
当m=n,i=1时,有(a+b)(a^(n-1)b+ab^(n-1))=(c+d)(c^(n-1)d+cd^(n-1)) => a^nb+a^2b^(n-1)+a^(n-1)b^2+ab^n=c^nd+c^2d^(n-1)+c^(n-1)d^2+cd^n---(6)
当m=n-1,i=0时,有(a^2+b^2)(a^(n-1)+b^(n-1))=(c^2+d^2)(c^(n-1)+d^(n-1)) => a^(n+1)+a^2b^(n-1)+a^(n-1)b^2+b^(n+1)=c^(n+1)+c^2d^(n-1)+c^(n-1)d^2+d^(n+1)---(7)
(5)(7)联立=> ab^n+a^nb-(a^2b^(n-1)+a^(n-1)b^2)=cd^n+c^nd-(c^2d^(n-1)+c^(n-1)d^2)---(8)
(6)(8)联立=> a^nb+ab^n=c^nd+cd^n---(9)
(5)(9)联立=> a^(n+1)+b^(n+1)=c^(n+1)+d^(n+1)
因此,对任意自然数n,都有a^n+b^n=c^n+d^n
当n=2003时,命题得证.