若n是自然数,且(n^3-1)/5是质数,求n的值

对n,5作带余数除法,设n=5k+r,k≥0,0≤r≤4.
n^3-1
=(5k+r)^3-1
=(5k+r-1)[25k^2+5k(2r+1)+r^2+r+1]
=5k[25k^2+5k(2r+1)+r^2+r+1]+(r-1)[25k^2+5k(2r+1)]+(r-1)(r^2+r+1)
要使n^3-1是5的倍数,(r-1)(r^2+r+1)必须是5的倍数,对于r=0,1,2,3,4只有r=1才行.
于是n=5k+1,将n代入
(n^3-1)/5=k(25k^2+15k+3),要使它是质数,k只能等于1,25k^2+15k+3=43的确是质数.
因此所求n值为n=5*1+1=6.
带入,算出为43,符合题意