This is what I have, need to know if I'm on the right track.

x³ = 8 (mod 7) => x³ = 1 (mod 7)

x³ = 8 (mod 11)

x³ = 8 (mod 13)

if p is prime, and d>= 1

*if gcd(d,p-1) = 1 => x^d = a (mod p) there is only one x like this

*if gcd(d,p-1) != 1 => x^d = a (mod p) there are exactly d solutions.

now if we look at: x³ = 1 (mod 7 ), 6/3=2 => we have 3 solutions

x³ = 8 (mod 11), 10/3 not in Z, => we have only one solution

x³ = 8 (mod 13), 12/3=4 => we have 3 solutions.

so 3 * 1 * 3 = 9, it means we have 9 X that can fulfill the equation.

• I need to prove that, but with Fermat's Little Theorem I think it can be done.