Can $3^x=2y^2-1$ be solved over the natural numbers? [closed]
$begingroup$
In the equation $3^x=2y^2-1$,
$x$, $y$ are natural numbers.
I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$)
but I even don't know if the number of solutions is infinite.
Is there a way to find the solution of this indeterminate equation?
diophantine-equations
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
$endgroup$
closed as off-topic by amWhy, user10354138, Saad, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user10354138, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 4 more comments
$begingroup$
In the equation $3^x=2y^2-1$,
$x$, $y$ are natural numbers.
I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$)
but I even don't know if the number of solutions is infinite.
Is there a way to find the solution of this indeterminate equation?
diophantine-equations
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
$endgroup$
closed as off-topic by amWhy, user10354138, Saad, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user10354138, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
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@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
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@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49
|
show 4 more comments
$begingroup$
In the equation $3^x=2y^2-1$,
$x$, $y$ are natural numbers.
I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$)
but I even don't know if the number of solutions is infinite.
Is there a way to find the solution of this indeterminate equation?
diophantine-equations
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
$endgroup$
In the equation $3^x=2y^2-1$,
$x$, $y$ are natural numbers.
I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$)
but I even don't know if the number of solutions is infinite.
Is there a way to find the solution of this indeterminate equation?
diophantine-equations
diophantine-equations
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
edited Dec 14 '18 at 7:56
eandpiandi
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
asked Dec 14 '18 at 6:47
eandpiandieandpiandi
272
272
closed as off-topic by amWhy, user10354138, Saad, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user10354138, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, user10354138, Saad, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, user10354138, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
$begingroup$
@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
$begingroup$
@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49
|
show 4 more comments
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
$begingroup$
@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
$begingroup$
@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
$begingroup$
@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
$begingroup$
@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49
|
show 4 more comments
2 Answers
2
active
oldest
votes
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Take both sides of the equation modulo $3$. If $xgeq 1$ then $2y^2equiv 1$ modulo $3$, but if $yequiv 0$, then $2y^2equiv 0$, and if $yequivpm 1$, $2y^2equiv 2(pm1)^2equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
$endgroup$
add a comment |
$begingroup$
You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.
And its solution is x=0 and y=1.
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take both sides of the equation modulo $3$. If $xgeq 1$ then $2y^2equiv 1$ modulo $3$, but if $yequiv 0$, then $2y^2equiv 0$, and if $yequivpm 1$, $2y^2equiv 2(pm1)^2equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
$endgroup$
add a comment |
$begingroup$
Take both sides of the equation modulo $3$. If $xgeq 1$ then $2y^2equiv 1$ modulo $3$, but if $yequiv 0$, then $2y^2equiv 0$, and if $yequivpm 1$, $2y^2equiv 2(pm1)^2equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
$endgroup$
add a comment |
$begingroup$
Take both sides of the equation modulo $3$. If $xgeq 1$ then $2y^2equiv 1$ modulo $3$, but if $yequiv 0$, then $2y^2equiv 0$, and if $yequivpm 1$, $2y^2equiv 2(pm1)^2equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
$endgroup$
Take both sides of the equation modulo $3$. If $xgeq 1$ then $2y^2equiv 1$ modulo $3$, but if $yequiv 0$, then $2y^2equiv 0$, and if $yequivpm 1$, $2y^2equiv 2(pm1)^2equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
answered Dec 14 '18 at 7:01
YiFanYiFan
4,3541627
4,3541627
add a comment |
add a comment |
$begingroup$
You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.
And its solution is x=0 and y=1.
$endgroup$
add a comment |
$begingroup$
You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.
And its solution is x=0 and y=1.
$endgroup$
add a comment |
$begingroup$
You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.
And its solution is x=0 and y=1.
$endgroup$
You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.
And its solution is x=0 and y=1.
answered Dec 20 '18 at 16:02
user628192
add a comment |
add a comment |
$begingroup$
"there was a mistake. I fixed the equation sry" What happens to YiFan's answer below, now offtopic? Bad practice...
$endgroup$
– Did
Dec 14 '18 at 7:38
$begingroup$
@Did Having seen the original state of the equation, it was probably more a typographical or less important error. (That is to say, it probably doesn't affect YiFan's answer in any huge way.)
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:41
$begingroup$
It unfortunately does, because the argument given in YiFan's answer no longer works. There is a solution to $2y^2 equiv -1 pmod 3$ so reducing modulo 3 does not lead to a contradiction (in fact there is a solution $pmod {3^k}$ for all $k$ by Hensel's lemma).
$endgroup$
– Tob Ernack
Dec 14 '18 at 7:44
$begingroup$
@EeveeTrainer It does, massively.
$endgroup$
– Did
Dec 14 '18 at 7:47
$begingroup$
Ah, I see... My bad then.
$endgroup$
– Eevee Trainer
Dec 14 '18 at 7:49