$2^{b}-1$ does not divide $2^{a}+1$
up vote
1
down vote
favorite
If $a$ and $b>2$ are any positive integers, then prove that $2^{a}+1$ is not divisible by $2^{b}-1$
Here's my attempt:
If $2^{b}-1mid 2^{a}+1$
then it is obvious that $a>b$.
Now ,By division algorithm
$$a=qb+r$$
We know that $$2^{b}equiv 1pmod{2^{b}-1}$$
Now raising to the power $q$ and then multiplying by $2^r$ we have,
$$2^{bq+r}equiv 2^{r}pmod{ 2^{b}-1}$$
But $$bq+r=a$$
Therefore,
$$2^{a}equiv 2^{r}pmod{ 2^{b}-1}$$
Adding one on both the sides
$$2^{a}+1equiv 2^{r}+1pmod{ 2^{b}-1}$$
As $r<b$,
$$2^{r}+1<2^{b}-1$$
Therefore, $$2^{r}+1notequiv 0 pmod{ 2^{b}-1}$$
Therefore, $2^{a}+1$ is not divisible by $2^{b}-1$ .
Can you show me where I went wrong?
divisibility
edited Nov 18 at 18:14
Bernard
116k637108
asked Nov 18 at 17:48
Samurai
987
add a comment |
up vote
1
down vote
favorite
If $a$ and $b>2$ are any positive integers, then prove that $2^{a}+1$ is not divisible by $2^{b}-1$
Here's my attempt:
If $2^{b}-1mid 2^{a}+1$
then it is obvious that $a>b$.
Now ,By division algorithm
$$a=qb+r$$
We know that $$2^{b}equiv 1pmod{2^{b}-1}$$
Now raising to the power $q$ and then multiplying by $2^r$ we have,
$$2^{bq+r}equiv 2^{r}pmod{ 2^{b}-1}$$
But $$bq+r=a$$
Therefore,
$$2^{a}equiv 2^{r}pmod{ 2^{b}-1}$$
Adding one on both the sides
$$2^{a}+1equiv 2^{r}+1pmod{ 2^{b}-1}$$
As $r<b$,
$$2^{r}+1<2^{b}-1$$
Therefore, $$2^{r}+1notequiv 0 pmod{ 2^{b}-1}$$
Therefore, $2^{a}+1$ is not divisible by $2^{b}-1$ .
Can you show me where I went wrong?
divisibility
edited Nov 18 at 18:14
Bernard
116k637108
asked Nov 18 at 17:48
Samurai
987
1
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
1
Is my proof correct?
– Samurai
Nov 18 at 18:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $a$ and $b>2$ are any positive integers, then prove that $2^{a}+1$ is not divisible by $2^{b}-1$
Here's my attempt:
If $2^{b}-1mid 2^{a}+1$
then it is obvious that $a>b$.
Now ,By division algorithm
$$a=qb+r$$
We know that $$2^{b}equiv 1pmod{2^{b}-1}$$
Now raising to the power $q$ and then multiplying by $2^r$ we have,
$$2^{bq+r}equiv 2^{r}pmod{ 2^{b}-1}$$
But $$bq+r=a$$
Therefore,
$$2^{a}equiv 2^{r}pmod{ 2^{b}-1}$$
Adding one on both the sides
$$2^{a}+1equiv 2^{r}+1pmod{ 2^{b}-1}$$
As $r<b$,
$$2^{r}+1<2^{b}-1$$
Therefore, $$2^{r}+1notequiv 0 pmod{ 2^{b}-1}$$
Therefore, $2^{a}+1$ is not divisible by $2^{b}-1$ .
Can you show me where I went wrong?
divisibility
edited Nov 18 at 18:14
Bernard
116k637108
asked Nov 18 at 17:48
Samurai
987
If $a$ and $b>2$ are any positive integers, then prove that $2^{a}+1$ is not divisible by $2^{b}-1$
Here's my attempt:
If $2^{b}-1mid 2^{a}+1$
then it is obvious that $a>b$.
Now ,By division algorithm
$$a=qb+r$$
We know that $$2^{b}equiv 1pmod{2^{b}-1}$$
Now raising to the power $q$ and then multiplying by $2^r$ we have,
$$2^{bq+r}equiv 2^{r}pmod{ 2^{b}-1}$$
But $$bq+r=a$$
Therefore,
$$2^{a}equiv 2^{r}pmod{ 2^{b}-1}$$
Adding one on both the sides
$$2^{a}+1equiv 2^{r}+1pmod{ 2^{b}-1}$$
As $r<b$,
$$2^{r}+1<2^{b}-1$$
Therefore, $$2^{r}+1notequiv 0 pmod{ 2^{b}-1}$$
Therefore, $2^{a}+1$ is not divisible by $2^{b}-1$ .
Can you show me where I went wrong?
divisibility
divisibility
edited Nov 18 at 18:14
Bernard
116k637108
asked Nov 18 at 17:48
Samurai
987
edited Nov 18 at 18:14
Bernard
116k637108
asked Nov 18 at 17:48
Samurai
987
edited Nov 18 at 18:14
Bernard
116k637108
edited Nov 18 at 18:14
Bernard
116k637108
edited Nov 18 at 18:14
Bernard
116k637108
116k637108
asked Nov 18 at 17:48
Samurai
987
asked Nov 18 at 17:48
Samurai
987
asked Nov 18 at 17:48
Samurai
987
987
1
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
1
Is my proof correct?
– Samurai
Nov 18 at 18:17
add a comment |
1
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
1
Is my proof correct?
– Samurai
Nov 18 at 18:17
1
1
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
1
1
Is my proof correct?
– Samurai
Nov 18 at 18:17
Is my proof correct?
– Samurai
Nov 18 at 18:17
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003859%2f2b-1-does-not-divide-2a1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
math.stackexchange.com/questions/495578/…
– lab bhattacharjee
Nov 18 at 18:14
1
Is my proof correct?
– Samurai
Nov 18 at 18:17