Let $p$ and $q$ are two prime numbers. Also, let us assume $q|(2^p-1)$. Then show that $(2^p-1,2^{q-1}-1)=2^{(p,q-1)}-1$. Note- $(p,q)$ denotes HCF of $p$ and $q$.
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I think you can use geometric series. – Jan 08 '16 at 18:31
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HCF? Highest common factor? (Here comes Freddy?) In the US we more frequently use GCD. Greatest common divisor. (Get Charlie down?) Regional biases, regional biases... gotta love them. (Oh, wait. No, I don't.) – fleablood Jan 08 '16 at 18:33
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It's true in general that $$\left (2^a-1,2^b-1 \right )=2^{(a,b)}-1$$ for $a$ , $b$ positive integers .
My favorite proof goes as follows :
Perform an induction on $a+b$ :
The base case $a+b=2$ so $a=b=1$ it's obvious .
Now assume that it's true for all $a$ and $b$ with $a+b \leq k$ .
Take some $a$ and $b$ with $a+b=k+1$ . Let also $d=\left (2^a-1,2^b-1 \right )$
Clearly $d$ is odd and also :
$$d \mid (2^a-1)-(2^b-1)=2^b(2^{a-b}-1)$$ (assuming $a \geq b$) so :
$$d \mid 2^{a-b}-1$$
This means that :
$$d \mid (2^b-1,2^{a-b}-1)$$
But because $b+a-b=a \leq k$ it folows that the RHS is $2^{(b,a-b)}-1$
Also it's clear that $(b,a-b)=(a,b)$ so :
$$d \mid 2^{(a,b)}-1$$
But also $$2^{(a,b)}-1 \mid d $$ because : $$2^{(a,b)}-1 \mid 2^a-1$$ and also :
$$2^{(a,b)}-1 \mid 2^b-1$$
This means that $d=2^{(a,b)}-1$ and the induction is finished :
NOTE : This is essentially equivalent with a proof using the Euclidean algorithm .
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@PITTALUGA Even $\left(a^m-b^m,a^n-b^n\right)=a^{(m,n)}-b^{(m,n)}$ for all $a,b,m,n\in\mathbb Z^+$ with $(a,b)=1$. See this answer. – user236182 Jan 08 '16 at 19:58