Show that if a and b are positive integers, then ( 2 a -1 ) mod ( 2 b − 1 ) = 2 a mod b − 1 . Use Exercise 36 to show that if a and b are positive integers, then gcd ( 2 a -1 , 2 b − 1 ) = 2 gcd ( a , b ) -1 . [Hint: Show that the remainders obtained when the Euclidean algorithm is used to compute gcd ( 2 a − 1 , 2 b -1 ) are of the from 2 r -1 , where r is a reminder arising when the Euclidean algorithm is used to find gcd ( a , b ) .]
Show that if a and b are positive integers, then ( 2 a -1 ) mod ( 2 b − 1 ) = 2 a mod b − 1 . Use Exercise 36 to show that if a and b are positive integers, then gcd ( 2 a -1 , 2 b − 1 ) = 2 gcd ( a , b ) -1 . [Hint: Show that the remainders obtained when the Euclidean algorithm is used to compute gcd ( 2 a − 1 , 2 b -1 ) are of the from 2 r -1 , where r is a reminder arising when the Euclidean algorithm is used to find gcd ( a , b ) .]
Solution Summary: The author explains how one can compute gcd using the Euclidean algorithm. If a and b are positive integers, the answer is 1 and the exponents involved in the continuing calculation are 2b and
Show that if a and b are positive integers, then
(
2
a
-1
)
mod
(
2
b
−
1
)
=
2
a
mod
b
−
1
.
Use Exercise 36 to show that if a and b are positive integers, then
gcd
(
2
a
-1
,
2
b
−
1
)
=
2
gcd
(
a
,
b
)
-1
.
[Hint: Show that the remainders obtained when the Euclidean algorithm is used to compute
gcd
(
2
a
−
1
,
2
b
-1
)
are of the from
2
r
-1
, where r is a reminder arising when the Euclidean algorithm is used to find
gcd
(
a
,
b
)
.]
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