Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$.

Question

Using Euclidean algorithm:

$$2^{2004}-1=4(2^{2002}-1)+3$$ $$2^{2002}-1=x\cdot 3+y$$

The solution manual says that $2^{2002}-1$has the remainder $0$ when divided by $3$, that is $y=0$ so GCD is $3$. But how do I find that remainder?

$2\equiv -1 \mod 3$. Hence $2^{2002}\equiv (-1)^{2002} \equiv 1 \mod 3$. — Leon Sot, Feb 02 '17 at 07:11
Also, @labbhattacharjee I would disagree that this question is a duplicate. Just because you can use that answer to answer this one, the linked question is much more general and the answers for it are overkill for this question. — Leon Sot, Feb 02 '17 at 07:15

score 1 · Answer 1 · answered Feb 02 '17 at 06:59

1

We have,

$2^{2002} - 1$

$= (2^4)^{500}.2^2 - 1$

Now $2^4 \equiv 1 (\mod 3)$

From above,

$= (1)^{500} .2^2 - 1$

$= 1.2^2 - 1$

$= 4 - 1 = 3$

Divisible by 3.

answered Feb 02 '17 at 06:59

Kanwaljit Singh

How is $(2^4)^{500}\cdot 2^2 - 1=1^{500}\cdot 2^2-1$? – lmc Feb 02 '17 at 07:08
$2^4 = 16 (mod 3) = 1$ – Kanwaljit Singh Feb 02 '17 at 07:15
Or simply you divide 16 by 3 them remainder 1 and power of 1 is any equals to 1. – Kanwaljit Singh Feb 02 '17 at 07:16

1 Answers1