Question

设a和n是任意自然数,求证a^(4n+1)-a能被30整除

Answer 1

a(a^4-1)整除a^(4n+1)-a

a(a^4-1)

=a(a^2-1)(a^2+1)

=a(a-1)(a+1)(a^2+1-5+5)

=a(a-1)(a+1)(a-2)(a+2)+5a(a-1)(a+1)

=(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)

上式为两个连乘积之和,第一个是5个连续整数之积,当然是30的倍数,后一个是5再乘以3个连续整数之积,也是30的倍数.