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The diophantine equation $5times 2^x-4=3^y-1$
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I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5times 2^x-4-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
diophantine-equations
asked Jul 21 at 18:27
Theory Nombre
1057
add a comment |Â
up vote
2
down vote
favorite
I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5times 2^x-4-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
diophantine-equations
asked Jul 21 at 18:27
Theory Nombre
1057
2
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5times 2^x-4-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
diophantine-equations
asked Jul 21 at 18:27
Theory Nombre
1057
I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5times 2^x-4-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
diophantine-equations
asked Jul 21 at 18:27
Theory Nombre
1057
asked Jul 21 at 18:27
Theory Nombre
1057
asked Jul 21 at 18:27
Theory Nombre
1057
asked Jul 21 at 18:27
Theory Nombre
1057
1057
2
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25
add a comment |Â
2
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25
2
2
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25
add a comment |Â
4 Answers
4
active
oldest
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up vote
1
down vote
With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$bigg[3^z-1=5(2^a)bigg] text and bigg[3^z+1=2^bbigg]$$
or
$$bigg[3^z+1=5(2^a)bigg] text and bigg[3^z-1=2^bbigg]$$
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
add a comment |Â
up vote
1
down vote
Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 cdot 16 = 81 - 1. $ Write this as
$5 cdot 16 cdot 2^x = 81 cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 cdot 2^x - 80 = 81 cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y geq 1.$
From $2^x equiv 1 pmod 81$ we get
$$ x equiv 0 pmod 54. $$
It follows that $2^x - 1$ is divisible by $2^54 - 1.$
$$ 2^54 - 1 = 3^4 cdot 7 cdot 19 cdot 73 cdot 87211 cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y equiv 1 pmod 262657$ we find
$$ y equiv 0 pmod 14592 $$ and especially
$$ y equiv 0 pmod 2^8. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y equiv 0 pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 cdot 5 cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) ; , $$
actually $x,y$ are both zero.
answered Jul 22 at 0:02
Will Jagy
97.1k594196
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
add a comment |Â
up vote
1
down vote
From $3^yequiv1$ mod $5$, we see that $4mid y$, so writing $y=4z$, we have
$$5cdot2^x-4=3^4z-1=(3^2z+1)(3^2z-1)$$
Since $3^2z+1equiv2$ mod $8$, we can only have $3^2z+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^2z-1=0$). Thus $3^2z+1=10$, so $z=1$ and thus $5cdot2^x-4=10cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
add a comment |Â
up vote
0
down vote
Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
answered Jul 21 at 18:31
DavidS
858
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$bigg[3^z-1=5(2^a)bigg] text and bigg[3^z+1=2^bbigg]$$
or
$$bigg[3^z+1=5(2^a)bigg] text and bigg[3^z-1=2^bbigg]$$
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
add a comment |Â
up vote
1
down vote
With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$bigg[3^z-1=5(2^a)bigg] text and bigg[3^z+1=2^bbigg]$$
or
$$bigg[3^z+1=5(2^a)bigg] text and bigg[3^z-1=2^bbigg]$$
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$bigg[3^z-1=5(2^a)bigg] text and bigg[3^z+1=2^bbigg]$$
or
$$bigg[3^z+1=5(2^a)bigg] text and bigg[3^z-1=2^bbigg]$$
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
With the two substitutions:
$$x-4=w$$
$$y=2z$$
We rearrange to:
$$5(2^w)=(3^z-1)(3^z+1)$$
So we want $z$ such that $$bigg[3^z-1=5(2^a)bigg] text and bigg[3^z+1=2^bbigg]$$
or
$$bigg[3^z+1=5(2^a)bigg] text and bigg[3^z-1=2^bbigg]$$
We can see $z=2$ (and thus $y=4$) satisfies the second line here. (which leads to $x=8$ as others conclude).
See if you can take it from here.
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
answered Jul 21 at 18:59
Rhys Hughes
3,8681227
3,8681227
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
add a comment |Â
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
Can we deduce that is a unique solution?
– Theory Nombre
Jul 21 at 19:02
add a comment |Â
up vote
1
down vote
Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 cdot 16 = 81 - 1. $ Write this as
$5 cdot 16 cdot 2^x = 81 cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 cdot 2^x - 80 = 81 cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y geq 1.$
From $2^x equiv 1 pmod 81$ we get
$$ x equiv 0 pmod 54. $$
It follows that $2^x - 1$ is divisible by $2^54 - 1.$
$$ 2^54 - 1 = 3^4 cdot 7 cdot 19 cdot 73 cdot 87211 cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y equiv 1 pmod 262657$ we find
$$ y equiv 0 pmod 14592 $$ and especially
$$ y equiv 0 pmod 2^8. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y equiv 0 pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 cdot 5 cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) ; , $$
actually $x,y$ are both zero.
answered Jul 22 at 0:02
Will Jagy
97.1k594196
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
add a comment |Â
up vote
1
down vote
Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 cdot 16 = 81 - 1. $ Write this as
$5 cdot 16 cdot 2^x = 81 cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 cdot 2^x - 80 = 81 cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y geq 1.$
From $2^x equiv 1 pmod 81$ we get
$$ x equiv 0 pmod 54. $$
It follows that $2^x - 1$ is divisible by $2^54 - 1.$
$$ 2^54 - 1 = 3^4 cdot 7 cdot 19 cdot 73 cdot 87211 cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y equiv 1 pmod 262657$ we find
$$ y equiv 0 pmod 14592 $$ and especially
$$ y equiv 0 pmod 2^8. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y equiv 0 pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 cdot 5 cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) ; , $$
actually $x,y$ are both zero.
answered Jul 22 at 0:02
Will Jagy
97.1k594196
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 cdot 16 = 81 - 1. $ Write this as
$5 cdot 16 cdot 2^x = 81 cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 cdot 2^x - 80 = 81 cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y geq 1.$
From $2^x equiv 1 pmod 81$ we get
$$ x equiv 0 pmod 54. $$
It follows that $2^x - 1$ is divisible by $2^54 - 1.$
$$ 2^54 - 1 = 3^4 cdot 7 cdot 19 cdot 73 cdot 87211 cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y equiv 1 pmod 262657$ we find
$$ y equiv 0 pmod 14592 $$ and especially
$$ y equiv 0 pmod 2^8. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y equiv 0 pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 cdot 5 cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) ; , $$
actually $x,y$ are both zero.
answered Jul 22 at 0:02
Will Jagy
97.1k594196
Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 cdot 16 = 81 - 1. $ Write this as
$5 cdot 16 cdot 2^x = 81 cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 cdot 2^x - 80 = 81 cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y geq 1.$
From $2^x equiv 1 pmod 81$ we get
$$ x equiv 0 pmod 54. $$
It follows that $2^x - 1$ is divisible by $2^54 - 1.$
$$ 2^54 - 1 = 3^4 cdot 7 cdot 19 cdot 73 cdot 87211 cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y equiv 1 pmod 262657$ we find
$$ y equiv 0 pmod 14592 $$ and especially
$$ y equiv 0 pmod 2^8. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y equiv 0 pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 cdot 5 cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) ; , $$
actually $x,y$ are both zero.
answered Jul 22 at 0:02
Will Jagy
97.1k594196
answered Jul 22 at 0:02
Will Jagy
97.1k594196
answered Jul 22 at 0:02
Will Jagy
97.1k594196
answered Jul 22 at 0:02
Will Jagy
97.1k594196
97.1k594196
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
add a comment |Â
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
Thank you very much. frankly, I just started this method like you.
– Theory Nombre
Jul 22 at 0:07
add a comment |Â
up vote
1
down vote
From $3^yequiv1$ mod $5$, we see that $4mid y$, so writing $y=4z$, we have
$$5cdot2^x-4=3^4z-1=(3^2z+1)(3^2z-1)$$
Since $3^2z+1equiv2$ mod $8$, we can only have $3^2z+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^2z-1=0$). Thus $3^2z+1=10$, so $z=1$ and thus $5cdot2^x-4=10cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
add a comment |Â
up vote
1
down vote
From $3^yequiv1$ mod $5$, we see that $4mid y$, so writing $y=4z$, we have
$$5cdot2^x-4=3^4z-1=(3^2z+1)(3^2z-1)$$
Since $3^2z+1equiv2$ mod $8$, we can only have $3^2z+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^2z-1=0$). Thus $3^2z+1=10$, so $z=1$ and thus $5cdot2^x-4=10cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
From $3^yequiv1$ mod $5$, we see that $4mid y$, so writing $y=4z$, we have
$$5cdot2^x-4=3^4z-1=(3^2z+1)(3^2z-1)$$
Since $3^2z+1equiv2$ mod $8$, we can only have $3^2z+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^2z-1=0$). Thus $3^2z+1=10$, so $z=1$ and thus $5cdot2^x-4=10cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
From $3^yequiv1$ mod $5$, we see that $4mid y$, so writing $y=4z$, we have
$$5cdot2^x-4=3^4z-1=(3^2z+1)(3^2z-1)$$
Since $3^2z+1equiv2$ mod $8$, we can only have $3^2z+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^2z-1=0$). Thus $3^2z+1=10$, so $z=1$ and thus $5cdot2^x-4=10cdot8$ implies $x-4=4$. We find $(x,y)=(8,4)$ as the only solution.
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
answered Jul 22 at 0:23
Barry Cipra
56.5k652118
56.5k652118
add a comment |Â
add a comment |Â
up vote
0
down vote
Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
answered Jul 21 at 18:31
DavidS
858
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
add a comment |Â
up vote
0
down vote
Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
answered Jul 21 at 18:31
DavidS
858
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
answered Jul 21 at 18:31
DavidS
858
Try $x=8$ and $y=4$. Not sure if there are any other integer solutions.
answered Jul 21 at 18:31
DavidS
858
answered Jul 21 at 18:31
DavidS
858
answered Jul 21 at 18:31
DavidS
858
answered Jul 21 at 18:31
DavidS
858
858
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
add a comment |Â
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Yes $(x,y)=(8,4)$ is a solution. How did you get the solution and is it unique?
– Theory Nombre
Jul 21 at 18:32
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
Guess and check. Sorry =)
– DavidS
Jul 21 at 18:34
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
This is more a comment, it's good you found a solution but now you need to solve the question in full generality.
– Daniel Buck
Jul 21 at 20:33
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
Can't comment since my rep is below 50 but you're right
– DavidS
Jul 21 at 21:05
add a comment |Â
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2
Catalan does not apply because $5cdot 2^x-4$ is not a perfect power.
– Peter
Jul 21 at 18:30
$(8/4)$ seems to be the only solution, but I do not know a proof yet. No further solution for $xle 10^5$
– Peter
Jul 21 at 18:33
maybe we have to solve an equation of form $3^yequiv 1( mod 2^x-4 )$??
– Theory Nombre
Jul 21 at 18:38
Usually, such equations are difficult to solve completely, but it is well possible that someone on this site finds a complete answer.
– Peter
Jul 21 at 18:40
compare various (elementary) answers at math.stackexchange.com/questions/1941354/… They take a few steps but are just the same observations repeated.. Also, this method is not much changed by the extra factor of $5$
– Will Jagy
Jul 21 at 19:25