Finding the HCF
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Find $(a^{2^m}+1, a^{2^n}+1)$ when a is odd and a,m,n are positive integers and m is not equal to n. I know that the hcf is a multiple of two but I can't prove that it is 2 which is the answer. Plz help.
number-theory elementary-number-theory divisibility greatest-common-divisor
asked Nov 8 at 5:27
Utkarsh Jha
326
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up vote
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down vote
favorite
Find $(a^{2^m}+1, a^{2^n}+1)$ when a is odd and a,m,n are positive integers and m is not equal to n. I know that the hcf is a multiple of two but I can't prove that it is 2 which is the answer. Plz help.
number-theory elementary-number-theory divisibility greatest-common-divisor
asked Nov 8 at 5:27
Utkarsh Jha
326
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find $(a^{2^m}+1, a^{2^n}+1)$ when a is odd and a,m,n are positive integers and m is not equal to n. I know that the hcf is a multiple of two but I can't prove that it is 2 which is the answer. Plz help.
number-theory elementary-number-theory divisibility greatest-common-divisor
asked Nov 8 at 5:27
Utkarsh Jha
326
Find $(a^{2^m}+1, a^{2^n}+1)$ when a is odd and a,m,n are positive integers and m is not equal to n. I know that the hcf is a multiple of two but I can't prove that it is 2 which is the answer. Plz help.
number-theory elementary-number-theory divisibility greatest-common-divisor
number-theory elementary-number-theory divisibility greatest-common-divisor
asked Nov 8 at 5:27
Utkarsh Jha
326
asked Nov 8 at 5:27
Utkarsh Jha
326
edited Nov 9 at 4:14
asked Nov 8 at 5:27
Utkarsh Jha
326
asked Nov 8 at 5:27
Utkarsh Jha
326
asked Nov 8 at 5:27
Utkarsh Jha
326
326
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55
add a comment |
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55
add a comment |
1 Answer
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a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
answered Nov 17 at 17:16
sirous
1,5391513
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
answered Nov 17 at 17:16
sirous
1,5391513
add a comment |
up vote
0
down vote
a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
answered Nov 17 at 17:16
sirous
1,5391513
add a comment |
up vote
0
down vote
up vote
0
down vote
a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
answered Nov 17 at 17:16
sirous
1,5391513
a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
answered Nov 17 at 17:16
sirous
1,5391513
answered Nov 17 at 17:16
sirous
1,5391513
answered Nov 17 at 17:16
sirous
1,5391513
answered Nov 17 at 17:16
sirous
1,5391513
1,5391513
add a comment |
add a comment |
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Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Nov 8 at 5:33
math.stackexchange.com/questions/123524/…
– lab bhattacharjee
Nov 8 at 5:51
Since $a$ is even, both the numbers are odd, so the HCF can't be $2$.
– B. Goddard
Nov 8 at 21:07
Special case of here where $,2nmid b-c,,$ so the gcd $ =(a+1,2) = 2 $ by $a$ odd.
– Bill Dubuque
Nov 12 at 2:55