To show congruence $3^8 \equiv -1 \pmod{17}$

Question

How to show that $3^8 ≡ -1 \pmod{17}$. I have tried by using value of $3^8$ but is there any other method available for solving when more higher powers are included ?

Answer 1

$$3^4=81\equiv-4\pmod{17}\implies3^8=(3^4)^2\equiv(-4)^2\equiv16\equiv-1$$

Answer 2

We have $3^{4} = 81 \equiv -4$ (mod $17$). So, $3^8 \equiv 16 \equiv -1$ (mod $17$).

Answer 3

Use Fermat's theorem:

If p is prime, and p doesn't divide a, then $a^{p-1} = 1 (mod p)$.

So $3^{16} = 1 \pmod{17}$.

Use that.

Answer 4

By Euler's Criterion and Quadratic Reciprocity: $$3^8\equiv\left(\frac{3}{17}\right)\equiv \left(\frac{17}{3}\right)\equiv \left(\frac{2}{3}\right)\equiv -1\pmod{17}$$