Using Fermat's Little Theorem for remainders

Question

Using Fermat's little theorem, we know that $$k^{p-2} \cdot k \equiv 1 \pmod p.$$

To find the multiplicative inverse of $6$ modulo $17$, we need to calculate $6^{15} \pmod {17}$. It's supposed to be all congruences hold modulo $17$.

$$6^{15} \equiv 6^8 \cdot 6^4 \cdot 6^2 \cdot 6 \equiv 16\cdot4\cdot2\cdot6 \equiv 3 \pmod {17}$$

I need help to understand the solution of $6^{15} \equiv 3 \pmod {17}$.

Thanks.

mod $17$, $6^2=36\equiv2$, so $6^4\equiv2^2=4$ and $6^8\equiv4^2=16$; therefore, $6^{15}\equiv6^8\cdot6^4\cdot6^2\cdot6\equiv16\cdot4\cdot2\cdot6=768\equiv3$ — J. W. Tanner, Jun 15 '20 at 22:18
Welcome to Math SE. The technique being used is the square and multiply method, such as described in this answer. — John Omielan, Jun 15 '20 at 22:19

score 1 · Accepted Answer · answered Jun 15 '20 at 22:18

$15 = 8+4+2+1$, hence $6^{15} = 6^8 \cdot 6^4\cdot 6^2\cdot 6^1$. Then:

$6^1 \equiv 6 \pmod{17}$

$6^2 = 36 \equiv 2 \pmod{17}$

$6^4 = (6^2)^2 \equiv 2^2 = 4 \pmod{17}$

$6^8 = (6^4)^2 \equiv 4^2 = 16 \pmod{17}$

Therefore $6^{15} \equiv 16 \cdot 4\cdot 2 \cdot 6 = 16\cdot 48 = (17-1)(51-3) \equiv 3 \pmod{17}$

Using Fermat's Little Theorem for remainders

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