Factoring $2^b-1$ out from $(1+2^a+…+2^(b-1)a)$
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It is well known that $2^ab-1=(2^a-1)(1+2^a+...+2^(b-1)a)=(2^b-1)(1+2^b+...+2^(a-1)b)$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^(b-1)a)$.
My question is simply how to factor out the factor $2^b-1$ from this expression.
elementary-number-theory factoring
asked Aug 18 at 4:17
Tejas Rao
265
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up vote
3
down vote
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It is well known that $2^ab-1=(2^a-1)(1+2^a+...+2^(b-1)a)=(2^b-1)(1+2^b+...+2^(a-1)b)$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^(b-1)a)$.
My question is simply how to factor out the factor $2^b-1$ from this expression.
elementary-number-theory factoring
asked Aug 18 at 4:17
Tejas Rao
265
1
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
It is well known that $2^ab-1=(2^a-1)(1+2^a+...+2^(b-1)a)=(2^b-1)(1+2^b+...+2^(a-1)b)$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^(b-1)a)$.
My question is simply how to factor out the factor $2^b-1$ from this expression.
elementary-number-theory factoring
asked Aug 18 at 4:17
Tejas Rao
265
It is well known that $2^ab-1=(2^a-1)(1+2^a+...+2^(b-1)a)=(2^b-1)(1+2^b+...+2^(a-1)b)$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^(b-1)a)$.
My question is simply how to factor out the factor $2^b-1$ from this expression.
elementary-number-theory factoring
asked Aug 18 at 4:17
Tejas Rao
265
asked Aug 18 at 4:17
Tejas Rao
265
asked Aug 18 at 4:17
Tejas Rao
265
asked Aug 18 at 4:17
Tejas Rao
265
265
1
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34
add a comment |Â
1
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34
1
1
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34
add a comment |Â
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1
Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear.
– Arthur
Aug 18 at 4:34