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 ?

You can have a look at this post: http://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand — Martin Sleziak, Oct 05 '15 at 07:13

score 3 · Accepted Answer · answered Oct 05 '15 at 05:42

3

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

answered Oct 05 '15 at 05:42

lab bhattacharjee

279,016

score 3 · Answer 2 · edited Nov 07 '15 at 14:51

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.

edited Nov 07 '15 at 14:51

Martin Sleziak

56,060

answered Oct 05 '15 at 05:47

fleablood

130,341

score 3 · Answer 3 · answered Oct 05 '15 at 13:28

3

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

answered Oct 05 '15 at 13:28

user236182

13,489

score 2 · Answer 4 · answered Oct 05 '15 at 05:43

2

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

answered Oct 05 '15 at 05:43

GAVD

7,436

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

4 Answers4

Linked