Question

设a,b,c是有理数,且有a+b+c=0,a^3+b^3+c^3=0,求证:对于任何奇数n都有a^n+b^n+c^n=0.

Answer 1

a+b+c=0→a=-(b+c)
0=a^3+b^3+c^3=[-(b+c)]^3+b^3+c^3=-(b^3+3b^2c+3bc^2+c^3)+b^3+c^3
=-3b^2c-3bc^2=-3bc(b+c)
故b=0
or c=0
or b+c=0
同理a+b+c=0→b=-(a+c)
→a=0
or c=0
or a+c=0
a+b+c=0→c=-(a+b)
→a=0
or b=0
or a+b=0
故a+b+c=0,a^3+b^3+c^3=0为a，b，c中一个为0，另两个呈相反数
or a=b=c=0
不妨设a=0，b=-c则
0=a^n+b^n+c^n=b^n+c^n=(-c)^n+c^n
显然正整数n应为奇数