Ord(a)=3(modp), so A 3 = 1 (MODP) or (A-1) (A2+A+1) = 0 (MODP).

A- 1≠0(modp) Otherwise, ord(a)= 1(modp), so a 2+a+1= 0 (MODP).

So (a+1) 6 = A6+6a5+15a4+20a3+15a2+6a+1= 21(A2+A+60

Only state the order (A+ 1) ≠ 1, 2,3 (MODP).

If ord(a+ 1)= 1(modp), then a=0(modp) is obviously impossible.

If ord(a+ 1)=2(modp), then A 2+2A+ 1 = A = 1 (MODP) is also wrong.

If order(A+ 1)= 3(modp), then a 3+3a2+3a1=1+3 (a 2+a+1-kloc-0/)+65438.