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Prove the following: $$ gcd(n^a-1, n^b-1) = n^{gcd(a, b)}-1 $$ I am not even sure where to start. I tried some stuff, but I always reach dead ends.

How should I go about proving this?

K A
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  • Hey good, you tried some stuff. Then you will have no problem writing the stuff into the question. Nice and easy. How did you try stuff without knowing where to get started, btw? No matter, I guess we'll see when we see the stuff. – rschwieb Apr 02 '18 at 16:36

2 Answers2

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Hint:

Let $q$ be the quotient if the Euclidean division of $a$ by $b$, $r=a\bmod b$ its remainder. Prove the remainder in the division of $n^a-1$ by $n^b-1$ is $n^r-1$, in other words: $$(n^a-1) \bmod (n^b-1)=n^{a\bmod b}-1.$$ Deduce that performing the Euclidean algorithm on $\; n^a-1\;$ and $\;n^b-1$ performs the Euclidean algorithm on $a$ and $b$ in parallel.

Bernard
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Can you factor $n^k - 1$? How many ways? Does it make any difference if $k$ is prime or composite.

What are the factors of $n^a - 1 = n^{\frac a{\gcd(a,b)}\gcd(a,b)}-1$ and what are the factors o $n^b - 1 = n^{\frac b{\gcd(a,b)}\gcd(a,b)} - 1$. Is $n^{\gcd(a,b)} -1$ a common factor? Is there any larger common factor?

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hint:

$n^k - 1 = (n-1)(n^{k-1} + ..... + n + 1)$.

And $n^{am} - 1 = (n^a)^m - 1 = (n^a - 1)(n^{a(m-1)} + ..... + n^a + 1)$.

So......

If $a = a'\gcd(a,b)$ and $b=b'\gcd(a,b)$ then $n^a - 1 = (n^{\gcd(a,b)})^{a'} - 1) = .....$ and $n^b -1 =(n^{\gcd(a,b)})^{b'} - 1)$.

So.....

fleablood
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