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Solution to Mordell's Equation $y^2=x^3+4$
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Here is my question:
Find all solutions to $y^2=x^3+4$.
My attempt:
Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.
Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce
$$y'^2=2x'^3+1.$$
However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.
diophantine-equations mordell-curves
asked Aug 4 at 2:15
blastzit
34219
add a comment |Â
up vote
1
down vote
favorite
Here is my question:
Find all solutions to $y^2=x^3+4$.
My attempt:
Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.
Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce
$$y'^2=2x'^3+1.$$
However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.
diophantine-equations mordell-curves
asked Aug 4 at 2:15
blastzit
34219
1
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is my question:
Find all solutions to $y^2=x^3+4$.
My attempt:
Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.
Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce
$$y'^2=2x'^3+1.$$
However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.
diophantine-equations mordell-curves
asked Aug 4 at 2:15
blastzit
34219
Here is my question:
Find all solutions to $y^2=x^3+4$.
My attempt:
Rewrite the equation as $(y-2)(y+2)=x^3$. Notice that if $y$ is odd, then $(y-2,y+2)=1$. Hence they are both cubes, but no cubes differ by $4$. Therefore $y$ is even and $x$ is even too.
Write $y=2y'$, then $4y'^2=x^3+4$. Write $x=2x'$, and we finally deduce
$$y'^2=2x'^3+1.$$
However, I cannot proceed after this step. Can anyone help? Thanks. The solution should be mainly done with elementary steps.
diophantine-equations mordell-curves
asked Aug 4 at 2:15
blastzit
34219
edited Aug 4 at 4:00
asked Aug 4 at 2:15
blastzit
34219
asked Aug 4 at 2:15
blastzit
34219
asked Aug 4 at 2:15
blastzit
34219
34219
1
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago
add a comment |Â
1
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago
1
1
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
If you are looking for integer points, then LMFDB curve 108.a2 (Cremona label 108a1) has the information that
$ x=0, y=pm2 $ are the only integer solutions. There is more information in the entry if you are curious, such as the Mordell-Weil group structure is $ mathbbZ/3mathbbZ. $
Here is a continuation from your attempt. From your last equation, and
changing notation, we get
$ (y_1-1)(y_1+1) = 2x_1^3. $ If $ y_1 $ is even we get a contradiction,
thus $ y_1 $ is odd. But now, if $ x_1 $ is odd we get a contradiction,
thus $ x_1 $ is even. Thus, let $ y_1 = 2y_2 + 1, x_1 = 2x_2. $
The equation is now $ 2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3. $
This reduces to $ y_2(y_2+1)/2 = 2x_2^3 $ where the left side is a
triangular number which is twice a cubic number. This is close to but
not the same as a
cubic triangular number.
answered Aug 4 at 3:39
Somos
10.9k1831
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you are looking for integer points, then LMFDB curve 108.a2 (Cremona label 108a1) has the information that
$ x=0, y=pm2 $ are the only integer solutions. There is more information in the entry if you are curious, such as the Mordell-Weil group structure is $ mathbbZ/3mathbbZ. $
Here is a continuation from your attempt. From your last equation, and
changing notation, we get
$ (y_1-1)(y_1+1) = 2x_1^3. $ If $ y_1 $ is even we get a contradiction,
thus $ y_1 $ is odd. But now, if $ x_1 $ is odd we get a contradiction,
thus $ x_1 $ is even. Thus, let $ y_1 = 2y_2 + 1, x_1 = 2x_2. $
The equation is now $ 2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3. $
This reduces to $ y_2(y_2+1)/2 = 2x_2^3 $ where the left side is a
triangular number which is twice a cubic number. This is close to but
not the same as a
cubic triangular number.
answered Aug 4 at 3:39
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
If you are looking for integer points, then LMFDB curve 108.a2 (Cremona label 108a1) has the information that
$ x=0, y=pm2 $ are the only integer solutions. There is more information in the entry if you are curious, such as the Mordell-Weil group structure is $ mathbbZ/3mathbbZ. $
Here is a continuation from your attempt. From your last equation, and
changing notation, we get
$ (y_1-1)(y_1+1) = 2x_1^3. $ If $ y_1 $ is even we get a contradiction,
thus $ y_1 $ is odd. But now, if $ x_1 $ is odd we get a contradiction,
thus $ x_1 $ is even. Thus, let $ y_1 = 2y_2 + 1, x_1 = 2x_2. $
The equation is now $ 2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3. $
This reduces to $ y_2(y_2+1)/2 = 2x_2^3 $ where the left side is a
triangular number which is twice a cubic number. This is close to but
not the same as a
cubic triangular number.
answered Aug 4 at 3:39
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you are looking for integer points, then LMFDB curve 108.a2 (Cremona label 108a1) has the information that
$ x=0, y=pm2 $ are the only integer solutions. There is more information in the entry if you are curious, such as the Mordell-Weil group structure is $ mathbbZ/3mathbbZ. $
Here is a continuation from your attempt. From your last equation, and
changing notation, we get
$ (y_1-1)(y_1+1) = 2x_1^3. $ If $ y_1 $ is even we get a contradiction,
thus $ y_1 $ is odd. But now, if $ x_1 $ is odd we get a contradiction,
thus $ x_1 $ is even. Thus, let $ y_1 = 2y_2 + 1, x_1 = 2x_2. $
The equation is now $ 2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3. $
This reduces to $ y_2(y_2+1)/2 = 2x_2^3 $ where the left side is a
triangular number which is twice a cubic number. This is close to but
not the same as a
cubic triangular number.
answered Aug 4 at 3:39
Somos
10.9k1831
If you are looking for integer points, then LMFDB curve 108.a2 (Cremona label 108a1) has the information that
$ x=0, y=pm2 $ are the only integer solutions. There is more information in the entry if you are curious, such as the Mordell-Weil group structure is $ mathbbZ/3mathbbZ. $
Here is a continuation from your attempt. From your last equation, and
changing notation, we get
$ (y_1-1)(y_1+1) = 2x_1^3. $ If $ y_1 $ is even we get a contradiction,
thus $ y_1 $ is odd. But now, if $ x_1 $ is odd we get a contradiction,
thus $ x_1 $ is even. Thus, let $ y_1 = 2y_2 + 1, x_1 = 2x_2. $
The equation is now $ 2y_2(2y_2+2) = 8y_2(y_2+1)/2 = 16x_2^3. $
This reduces to $ y_2(y_2+1)/2 = 2x_2^3 $ where the left side is a
triangular number which is twice a cubic number. This is close to but
not the same as a
cubic triangular number.
answered Aug 4 at 3:39
Somos
10.9k1831
edited 2 hours ago
answered Aug 4 at 3:39
Somos
10.9k1831
answered Aug 4 at 3:39
Somos
10.9k1831
answered Aug 4 at 3:39
Somos
10.9k1831
10.9k1831
add a comment |Â
add a comment |Â
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1
We need all rational solutions? Or rather the "simpler situation" of finding all integer solutions?
– dan_fulea
Aug 4 at 2:33
@dan_fulea Only integer solutions.
– blastzit
Aug 4 at 2:47
You can try and continue along the same path: $(y'-1)(y'+1)=2x'^3$. Now, $gcd(y'-1, y'+1)=2$, so (apart from the trivial $x'=0$ solution) either $y'-1=2a^3$, $y'+1=b^3$, or $y'-1=b^3$, $y'+1=2a^3$ (with $a$ odd and $b$ even in either case, and $gcd(a,b)=1$); this gives you equations of the form $4b'^3-a^3=pm1$ (where $b=2b'$), though there's no immediate argument that comes to mind for tackling these.
– Steven Stadnicki
2 days ago