Solutions for $2^n+1=p^q$
up vote
0
down vote
favorite
I have a problem I can’t solve, please help!
Find all positive integer triples $(n,p,q)$ satisfying $2^n+1=p^q$, where $p,q>1$.
There is a similar problem I can solve: Prove that it is not possible that $2^n-1=p^q$, if $p,q>1$.
My solution:
We need to prove that $2^n=p^q+1$ is not possible. Note that $p$ is an odd number and if you check $mod 4$ then you find that $q$ is also an odd number. Then $2^n=p^q+1=p^q+1^q=(p+1)(p^{q-1}-p^{q-2}+dots -p+1)$. Note that $2^n$ doesn’t have an odd divisor $>1$, but since $(p^{q-1}-p^{q-2}+dots -p+1)$ is an odd number $>1$, contradiction.
number-theory elementary-number-theory diophantine-equations
|
show 1 more comment
up vote
0
down vote
favorite
I have a problem I can’t solve, please help!
Find all positive integer triples $(n,p,q)$ satisfying $2^n+1=p^q$, where $p,q>1$.
There is a similar problem I can solve: Prove that it is not possible that $2^n-1=p^q$, if $p,q>1$.
My solution:
We need to prove that $2^n=p^q+1$ is not possible. Note that $p$ is an odd number and if you check $mod 4$ then you find that $q$ is also an odd number. Then $2^n=p^q+1=p^q+1^q=(p+1)(p^{q-1}-p^{q-2}+dots -p+1)$. Note that $2^n$ doesn’t have an odd divisor $>1$, but since $(p^{q-1}-p^{q-2}+dots -p+1)$ is an odd number $>1$, contradiction.
number-theory elementary-number-theory diophantine-equations
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
1
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a problem I can’t solve, please help!
Find all positive integer triples $(n,p,q)$ satisfying $2^n+1=p^q$, where $p,q>1$.
There is a similar problem I can solve: Prove that it is not possible that $2^n-1=p^q$, if $p,q>1$.
My solution:
We need to prove that $2^n=p^q+1$ is not possible. Note that $p$ is an odd number and if you check $mod 4$ then you find that $q$ is also an odd number. Then $2^n=p^q+1=p^q+1^q=(p+1)(p^{q-1}-p^{q-2}+dots -p+1)$. Note that $2^n$ doesn’t have an odd divisor $>1$, but since $(p^{q-1}-p^{q-2}+dots -p+1)$ is an odd number $>1$, contradiction.
number-theory elementary-number-theory diophantine-equations
I have a problem I can’t solve, please help!
Find all positive integer triples $(n,p,q)$ satisfying $2^n+1=p^q$, where $p,q>1$.
There is a similar problem I can solve: Prove that it is not possible that $2^n-1=p^q$, if $p,q>1$.
My solution:
We need to prove that $2^n=p^q+1$ is not possible. Note that $p$ is an odd number and if you check $mod 4$ then you find that $q$ is also an odd number. Then $2^n=p^q+1=p^q+1^q=(p+1)(p^{q-1}-p^{q-2}+dots -p+1)$. Note that $2^n$ doesn’t have an odd divisor $>1$, but since $(p^{q-1}-p^{q-2}+dots -p+1)$ is an odd number $>1$, contradiction.
number-theory elementary-number-theory diophantine-equations
number-theory elementary-number-theory diophantine-equations
edited Nov 16 at 23:41
Servaes
20.6k33789
20.6k33789
asked Nov 14 at 14:45
Ti Tu Lea
284
284
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
1
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53
|
show 1 more comment
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
1
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
1
1
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53
|
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have
$$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+cdots + p+1)$$
Hence $p$ is odd, $p^{q-1}+p^{q-2}+cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then
$$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$
so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.
The second case the proof you wrote works perfectly.
add a comment |
up vote
2
down vote
These are both special cases of Mihăilescu's theorem, which says that the only solution to
$$x^a-y^b=1,$$
in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.
add a comment |
up vote
1
down vote
As $p-1mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $kge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).
If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution
$$ 2^3+1=3^2.$$
If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(ldots)2^{2k}-1=bigl(q+(ldots)2^kbigr)cdot 2^k$$
is a proper odd multiple of $2^k$, contradiction.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have
$$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+cdots + p+1)$$
Hence $p$ is odd, $p^{q-1}+p^{q-2}+cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then
$$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$
so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.
The second case the proof you wrote works perfectly.
add a comment |
up vote
1
down vote
accepted
I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have
$$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+cdots + p+1)$$
Hence $p$ is odd, $p^{q-1}+p^{q-2}+cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then
$$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$
so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.
The second case the proof you wrote works perfectly.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have
$$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+cdots + p+1)$$
Hence $p$ is odd, $p^{q-1}+p^{q-2}+cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then
$$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$
so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.
The second case the proof you wrote works perfectly.
I try to explain an elementary proof. First, we consider the equation $2^n=p^q-1$. We have
$$2^n=p^q-1=(p-1)(p^{q-1}+p^{q-2}+cdots + p+1)$$
Hence $p$ is odd, $p^{q-1}+p^{q-2}+cdots + p+1$ is even, and hence $q$ must be even. But if $q=2s$, then
$$2^n=p^{2s}-1=(p^s-1)(p^s+1)$$
so $p^s=2^r+1=2^t-1$ for some $r,t$ with $r+s=n$, hence $2^t=2^r+2=2(2^{r-1}+1)$, so $2^{r-1}$ is odd, which only happens when $r=1$, so $t=2$, $n=3$ and hence $p=3$, $q=2$.
The second case the proof you wrote works perfectly.
answered Nov 14 at 15:17
xarles
1,48079
1,48079
add a comment |
add a comment |
up vote
2
down vote
These are both special cases of Mihăilescu's theorem, which says that the only solution to
$$x^a-y^b=1,$$
in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.
add a comment |
up vote
2
down vote
These are both special cases of Mihăilescu's theorem, which says that the only solution to
$$x^a-y^b=1,$$
in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.
add a comment |
up vote
2
down vote
up vote
2
down vote
These are both special cases of Mihăilescu's theorem, which says that the only solution to
$$x^a-y^b=1,$$
in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.
These are both special cases of Mihăilescu's theorem, which says that the only solution to
$$x^a-y^b=1,$$
in integers $a,b>1$ and $x,y>0$ is $(a,b,x,y)=(2,3,3,2)$.
answered Nov 14 at 14:56
Servaes
20.6k33789
20.6k33789
add a comment |
add a comment |
up vote
1
down vote
As $p-1mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $kge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).
If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution
$$ 2^3+1=3^2.$$
If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(ldots)2^{2k}-1=bigl(q+(ldots)2^kbigr)cdot 2^k$$
is a proper odd multiple of $2^k$, contradiction.
add a comment |
up vote
1
down vote
As $p-1mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $kge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).
If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution
$$ 2^3+1=3^2.$$
If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(ldots)2^{2k}-1=bigl(q+(ldots)2^kbigr)cdot 2^k$$
is a proper odd multiple of $2^k$, contradiction.
add a comment |
up vote
1
down vote
up vote
1
down vote
As $p-1mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $kge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).
If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution
$$ 2^3+1=3^2.$$
If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(ldots)2^{2k}-1=bigl(q+(ldots)2^kbigr)cdot 2^k$$
is a proper odd multiple of $2^k$, contradiction.
As $p-1mid p^q-1=2^n$, we see that $p$ must be one more than a power of $2$, so $p=2^k+1$ (with $kge1$ because $p=2^0+1$ leads only to $n=0$, $q=1$).
If $q$ is even, then $2^n=p^q-1=(p^{q/2}-1)(p^{q/2}+1)$, so $p^{q/2}pm1$ must be powers of $2$, making $p^{q/2}=3$, $p=3$, $q=2$. We have found the solution
$$ 2^3+1=3^2.$$
If $q$ is odd, then $$2^n=q^p-1=(1+2^k)^q-1 =1+q2^k+(ldots)2^{2k}-1=bigl(q+(ldots)2^kbigr)cdot 2^k$$
is a proper odd multiple of $2^k$, contradiction.
answered Nov 14 at 15:10
Hagen von Eitzen
274k21266494
274k21266494
add a comment |
add a comment |
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%2f2998341%2fsolutions-for-2n1-pq%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
For $2^n=p^q+1$, isnt n=2, p = 3 and q = 1 a solution?
– QuIcKmAtHs
Nov 14 at 14:47
Right, I missed that $p,q>1$
– Ti Tu Lea
Nov 14 at 14:48
Yes, that is much more logical now.
– QuIcKmAtHs
Nov 14 at 14:48
I appreciate your effort in solving this question, but the equation to be solved in the solution part doesn't correspond to the one in the problem statement.
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 14:48
1
Isn't $2^3+1=3^2$? So there is a solution possible.
– 5xum
Nov 14 at 14:53