Modulo Problem, Fermat's little theorem

Question

Find the value of the unique integer x satisfying $O \le x \le 17$ for which $$ 4^{1024000000002} \equiv x\pmod{17} $$ I think this is related to Fermat's little theorem. I'm knowledgeable with the Chinese remainder theorem and just need some advice on solving this.

Answer 1

You can indeed use Fermat's little theorem. Since 17 is prime and 4 is relatively prime to 17, we know that $a^{17-1}=a^{16}\equiv 1\pmod{17}$. Now since $$ 1024000000002 = 16\cdot64000000000+2 $$ we'll have $$ 4^{1024000000002}=4^{16\cdot64000000000+2}=(4^{16})^{64000000000}(4^2)\equiv(1)(16)\pmod{17} $$

Answer 2

HINT:

As $4^2=16\equiv-1\pmod {17}$ and $a\equiv b\pmod m\implies a^n\equiv b^n\pmod m$ for integer $n\ge0,$

$4^{1024000000002}=(4^2)^{512000000001}\equiv(-1)^{512000000001}\pmod{17}\equiv-1\equiv16$