Question

mn是奇偶性相同的自然数证(a^m+b^m)(a^n+b^n)

Answer 1

(a^m+b^m)(a^n+b^n)-2(a^(m+n)+b^(m+n))

=a^(m+n)+a^n*b^m+a^m*b^n+b^(m+n)-2a^(m+n)-2b^(m+n)

=-a^m*a^n-b^m*b^n+a^n*b^m+a^m*b^n

=-a^m(a^n-b^n)+b^m(a^n-b^n)

=-(a^m-b^m)(a^n-b^n)

当a>b时：a^m-b^m>0;a^n-b^n>0,等式