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

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?

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    What is $2^{2002}-1 \mod 3$? – 2017-02-02
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    @LeonSot That's my question, yes. – 2017-02-02
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    $2\equiv -1 \mod 3$. Hence $2^{2002}\equiv (-1)^{2002} \equiv 1 \mod 3$. – 2017-02-02
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    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. – 2017-02-02

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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.

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    How is $(2^4)^{500}\cdot 2^2 - 1=1^{500}\cdot 2^2-1$? – 2017-02-02
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    $2^4 = 16 (mod 3) = 1$ – 2017-02-02
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    Or simply you divide 16 by 3 them remainder 1 and power of 1 is any equals to 1. – 2017-02-02