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Parameterization to the Diophantine equation $x^2+4y^2=5z^2$ [closed]
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I'd like to ask how to obtain all integer solutions to the Diophantine equation $x^2+4y^2=5z^2$ using parameterization? Thanks.
number-theory diophantine-equations parametrization
asked Jul 21 at 17:24
Hang Wu
658
closed as off-topic by user223391, José Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel Jul 22 at 3:05
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I'd like to ask how to obtain all integer solutions to the Diophantine equation $x^2+4y^2=5z^2$ using parameterization? Thanks.
number-theory diophantine-equations parametrization
asked Jul 21 at 17:24
Hang Wu
658
closed as off-topic by user223391, José Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel Jul 22 at 3:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РCommunity, Jos̩ Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel
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math.stackexchange.com/questions/2773097/…
– individ
Jul 21 at 17:51
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up vote
1
down vote
favorite
I'd like to ask how to obtain all integer solutions to the Diophantine equation $x^2+4y^2=5z^2$ using parameterization? Thanks.
number-theory diophantine-equations parametrization
asked Jul 21 at 17:24
Hang Wu
658
I'd like to ask how to obtain all integer solutions to the Diophantine equation $x^2+4y^2=5z^2$ using parameterization? Thanks.
number-theory diophantine-equations parametrization
asked Jul 21 at 17:24
Hang Wu
658
asked Jul 21 at 17:24
Hang Wu
658
asked Jul 21 at 17:24
Hang Wu
658
asked Jul 21 at 17:24
Hang Wu
658
658
closed as off-topic by user223391, José Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel Jul 22 at 3:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РCommunity, Jos̩ Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel
closed as off-topic by user223391, José Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel Jul 22 at 3:05
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РCommunity, Jos̩ Carlos Santos, Adrian Keister, Leucippus, Parcly Taxel
1
math.stackexchange.com/questions/2773097/…
– individ
Jul 21 at 17:51
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math.stackexchange.com/questions/2773097/…
– individ
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math.stackexchange.com/questions/2773097/…
– individ
Jul 21 at 17:51
math.stackexchange.com/questions/2773097/…
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3 Answers
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Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $xequiv pm ypmod5$. We may assume that $xequiv ypmod5$ (otherwise, just flip the sign of $y$). Now, write
$$z^2=fracx^2+4y^25=left(fracx+4y5right)^2+left(frac2x-2y5right)^2,.$$
That is, using the knowledge from Pythagorian triples, we get that
$$fracx+4y5=m^2-n^2,,,,frac2x-2y5=2mn,,text and z=m^2+n^2$$
for some coprime $m,ninmathbbZ$ with different parity. Thus,
$$x=m^2+4mn-n^2,,,,y=m^2-mn-n^2,,text and z=m^2+n^2,.$$
Note that, if you set $m:=u$ and $n:=u+v$, then you will get the same thing as in Will Jagy's answer (up to sign changes).
answered Jul 21 at 18:08
Batominovski
23.3k22777
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
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The methods of Fricke and Klein (1897) show that a complete parametrization is given by a finite number of this type:
$$ x = 4 u^2 + 2 uv - v^2 ; , ; ; y = u^2 + 3 uv + v^2 ; , ; ; z = 2u^2 + 2 uv + v^2 ; , ; ; $$
where we do not restrict $u,v$ to positive, then freely use absolute values for $x,y,z.$ I will need to check whether this one recipe gives them all. If you want $x,y,z$ to have a common factor, simply multiply them all by some number.
Oh, note that $z$ comes out positive regardless, and we can bound $u,v,$ if we are bounding $z, ;$ since $2u^2 + 2uv+v^2 = u^2 + (u+v)^2 ; .$
Note that stereographic projection around a single point easily gives all rational solutions. However, there is then the problem of finding the primitive integer solutions.
One is enough. The above is sufficient. Oh, I was forgetting, might as well take $u geq 0$ to reduce repetition.
Sat Jul 21 11:06:37 PDT 2018
|z| x y z u v
1 -1 1 1 0 1
1 -1 1 1 0 -1
1 1 -1 1 1 -1
2 4 1 2 1 0
2 -4 -1 2 1 -2
5 -11 1 5 1 -3
5 11 -1 5 2 -1
10 4 11 10 1 2
10 -4 -11 10 3 -4
13 19 11 13 2 1
13 -19 -11 13 3 -5
13 -29 -1 13 2 -5
13 29 1 13 3 -1
17 1 19 17 1 3
17 -1 -19 17 4 -5
17 -31 11 17 1 -5
17 31 -11 17 4 -3
25 41 19 25 3 1
25 -41 -19 25 4 -7
26 -4 29 26 1 4
26 4 -29 26 5 -6
26 -44 19 26 1 -6
26 44 -19 26 5 -4
29 19 31 29 2 3
29 -19 -31 29 5 -7
29 -61 11 29 2 -7
29 61 -11 29 5 -3
34 44 31 34 3 2
34 -44 -31 34 5 -8
34 -76 1 34 3 -8
34 76 -1 34 5 -2
37 -11 41 37 1 5
37 11 -41 37 6 -7
37 -59 29 37 1 -7
37 59 -29 37 6 -5
41 71 29 41 4 1
41 -71 -29 41 5 -9
41 -89 -11 41 4 -9
41 89 11 41 5 -1
50 -76 41 50 1 -8
50 76 -41 50 7 -6
53 -101 31 53 2 -9
53 101 -31 53 7 -5
53 11 59 53 2 5
53 -11 -59 53 7 -9
58 -124 19 58 3 -10
58 124 -19 58 7 -4
58 44 61 58 3 4
58 -44 -61 58 7 -10
61 109 41 61 5 1
61 -109 -41 61 6 -11
61 -131 -19 61 5 -11
61 131 19 61 6 -1
65 -31 71 65 1 7
65 31 -71 65 8 -9
65 79 61 65 4 3
65 -79 -61 65 7 -11
73 -151 31 73 3 -11
73 151 -31 73 8 -5
73 41 79 73 3 5
73 -41 -79 73 8 -11
74 116 59 74 5 2
74 -116 -59 74 7 -12
74 -164 -11 74 5 -12
74 164 11 74 7 -2
82 -116 71 82 1 -10
82 116 -71 82 9 -8
82 -44 89 82 1 8
82 44 -89 82 9 -10
85 -149 59 85 2 -11
85 149 -59 85 9 -7
85 -181 -29 85 6 -13
85 181 29 85 7 -1
89 121 79 89 5 3
89 -121 -79 89 8 -13
89 -199 -1 89 5 -13
89 199 1 89 8 -3
97 -209 29 97 4 -13
97 209 -29 97 9 -5
97 79 101 97 4 5
97 -79 -101 97 9 -13
101 139 -89 101 10 -9
101 -139 89 101 1 -11
101 59 -109 101 10 -11
101 -59 109 101 1 9
106 124 101 106 5 4
106 -124 -101 106 9 -14
106 -236 11 106 5 -14
106 236 -11 106 9 -4
109 211 -61 109 10 -7
109 -211 61 109 3 -13
109 -29 -121 109 10 -13
109 29 121 109 3 7
113 209 71 113 7 1
113 -209 -71 113 8 -15
113 -239 -41 113 7 -15
113 239 41 113 8 -1
122 164 -109 122 11 -10
122 -164 109 122 1 -12
122 -76 131 122 1 10
122 76 -131 122 11 -12
125 29 -139 125 11 -13
125 -29 139 125 2 9
130 244 -79 130 11 -8
130 -244 79 130 3 -14
130 -284 -31 130 7 -16
130 284 31 130 9 -2
137 281 -61 137 11 -7
137 -281 61 137 4 -15
137 -71 -149 137 11 -15
137 71 149 137 4 7
145 -191 131 145 1 -13
145 191 -131 145 12 -11
145 271 89 145 8 1
145 -271 -89 145 9 -17
146 -124 -151 146 11 -16
146 124 151 146 5 6
146 316 -41 146 11 -6
146 -316 41 146 5 -16
149 -229 -121 149 10 -17
149 229 121 149 7 3
149 331 19 149 10 -3
149 -331 -19 149 7 -17
|z| x y z u v
=================================================
answered Jul 21 at 17:41
Will Jagy
97.1k594196
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
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Do you know how to do this for Pythagorean triples? That is, do you know how to find solutions to $x^2 + y^2 = z^2$? This is equivalent to finding rational points on the circle $X^2 + Y^2 = 1$, and then taking $(x/z,y/z) = (X,Y)$. The way to find rational points on the circle is as follows: for a point on the circle $(X,Y)$, consider the line from $(-1,0)$ to $(X,Y)$. If this line has rational slope, then the equation for where the line meets the circle has rational coefficients and has the rational root at $(-1,0)$. The rational root test then tells you that the other root must be rational, i.e. $(X,Y)$ is rational.
Conversely, if $(X,Y)$ is rational, then the line from $(-1,0)$ to $(X,Y)$ has rational slope as well. Therefore, finding rational points on the unit circle are equivalent to looking at where the lines of rational slope at $(-1,0)$ intersect the circle; this can be done using the quadratic formula.
The same can be used for your equation. Simply consider all lines of rational slope starting at some fixed rational point of $X^2 / 5 + 4 Y^2 / 5 = 1$, e.g. $(1,1)$.
answered Jul 21 at 17:42
Marcus M
8,1731847
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $xequiv pm ypmod5$. We may assume that $xequiv ypmod5$ (otherwise, just flip the sign of $y$). Now, write
$$z^2=fracx^2+4y^25=left(fracx+4y5right)^2+left(frac2x-2y5right)^2,.$$
That is, using the knowledge from Pythagorian triples, we get that
$$fracx+4y5=m^2-n^2,,,,frac2x-2y5=2mn,,text and z=m^2+n^2$$
for some coprime $m,ninmathbbZ$ with different parity. Thus,
$$x=m^2+4mn-n^2,,,,y=m^2-mn-n^2,,text and z=m^2+n^2,.$$
Note that, if you set $m:=u$ and $n:=u+v$, then you will get the same thing as in Will Jagy's answer (up to sign changes).
answered Jul 21 at 18:08
Batominovski
23.3k22777
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
add a comment |Â
up vote
2
down vote
Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $xequiv pm ypmod5$. We may assume that $xequiv ypmod5$ (otherwise, just flip the sign of $y$). Now, write
$$z^2=fracx^2+4y^25=left(fracx+4y5right)^2+left(frac2x-2y5right)^2,.$$
That is, using the knowledge from Pythagorian triples, we get that
$$fracx+4y5=m^2-n^2,,,,frac2x-2y5=2mn,,text and z=m^2+n^2$$
for some coprime $m,ninmathbbZ$ with different parity. Thus,
$$x=m^2+4mn-n^2,,,,y=m^2-mn-n^2,,text and z=m^2+n^2,.$$
Note that, if you set $m:=u$ and $n:=u+v$, then you will get the same thing as in Will Jagy's answer (up to sign changes).
answered Jul 21 at 18:08
Batominovski
23.3k22777
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $xequiv pm ypmod5$. We may assume that $xequiv ypmod5$ (otherwise, just flip the sign of $y$). Now, write
$$z^2=fracx^2+4y^25=left(fracx+4y5right)^2+left(frac2x-2y5right)^2,.$$
That is, using the knowledge from Pythagorian triples, we get that
$$fracx+4y5=m^2-n^2,,,,frac2x-2y5=2mn,,text and z=m^2+n^2$$
for some coprime $m,ninmathbbZ$ with different parity. Thus,
$$x=m^2+4mn-n^2,,,,y=m^2-mn-n^2,,text and z=m^2+n^2,.$$
Note that, if you set $m:=u$ and $n:=u+v$, then you will get the same thing as in Will Jagy's answer (up to sign changes).
answered Jul 21 at 18:08
Batominovski
23.3k22777
Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $xequiv pm ypmod5$. We may assume that $xequiv ypmod5$ (otherwise, just flip the sign of $y$). Now, write
$$z^2=fracx^2+4y^25=left(fracx+4y5right)^2+left(frac2x-2y5right)^2,.$$
That is, using the knowledge from Pythagorian triples, we get that
$$fracx+4y5=m^2-n^2,,,,frac2x-2y5=2mn,,text and z=m^2+n^2$$
for some coprime $m,ninmathbbZ$ with different parity. Thus,
$$x=m^2+4mn-n^2,,,,y=m^2-mn-n^2,,text and z=m^2+n^2,.$$
Note that, if you set $m:=u$ and $n:=u+v$, then you will get the same thing as in Will Jagy's answer (up to sign changes).
answered Jul 21 at 18:08
Batominovski
23.3k22777
edited Jul 21 at 18:34
answered Jul 21 at 18:08
Batominovski
23.3k22777
answered Jul 21 at 18:08
Batominovski
23.3k22777
answered Jul 21 at 18:08
Batominovski
23.3k22777
23.3k22777
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
add a comment |Â
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
This is better. My Fricke Klein program gives me a big list of possibilities (as 3 by 3 matrices of integers). I was not patient enough to look in the middle for this more attractive parametrization.
– Will Jagy
Jul 21 at 18:26
add a comment |Â
up vote
1
down vote
The methods of Fricke and Klein (1897) show that a complete parametrization is given by a finite number of this type:
$$ x = 4 u^2 + 2 uv - v^2 ; , ; ; y = u^2 + 3 uv + v^2 ; , ; ; z = 2u^2 + 2 uv + v^2 ; , ; ; $$
where we do not restrict $u,v$ to positive, then freely use absolute values for $x,y,z.$ I will need to check whether this one recipe gives them all. If you want $x,y,z$ to have a common factor, simply multiply them all by some number.
Oh, note that $z$ comes out positive regardless, and we can bound $u,v,$ if we are bounding $z, ;$ since $2u^2 + 2uv+v^2 = u^2 + (u+v)^2 ; .$
Note that stereographic projection around a single point easily gives all rational solutions. However, there is then the problem of finding the primitive integer solutions.
One is enough. The above is sufficient. Oh, I was forgetting, might as well take $u geq 0$ to reduce repetition.
Sat Jul 21 11:06:37 PDT 2018
|z| x y z u v
1 -1 1 1 0 1
1 -1 1 1 0 -1
1 1 -1 1 1 -1
2 4 1 2 1 0
2 -4 -1 2 1 -2
5 -11 1 5 1 -3
5 11 -1 5 2 -1
10 4 11 10 1 2
10 -4 -11 10 3 -4
13 19 11 13 2 1
13 -19 -11 13 3 -5
13 -29 -1 13 2 -5
13 29 1 13 3 -1
17 1 19 17 1 3
17 -1 -19 17 4 -5
17 -31 11 17 1 -5
17 31 -11 17 4 -3
25 41 19 25 3 1
25 -41 -19 25 4 -7
26 -4 29 26 1 4
26 4 -29 26 5 -6
26 -44 19 26 1 -6
26 44 -19 26 5 -4
29 19 31 29 2 3
29 -19 -31 29 5 -7
29 -61 11 29 2 -7
29 61 -11 29 5 -3
34 44 31 34 3 2
34 -44 -31 34 5 -8
34 -76 1 34 3 -8
34 76 -1 34 5 -2
37 -11 41 37 1 5
37 11 -41 37 6 -7
37 -59 29 37 1 -7
37 59 -29 37 6 -5
41 71 29 41 4 1
41 -71 -29 41 5 -9
41 -89 -11 41 4 -9
41 89 11 41 5 -1
50 -76 41 50 1 -8
50 76 -41 50 7 -6
53 -101 31 53 2 -9
53 101 -31 53 7 -5
53 11 59 53 2 5
53 -11 -59 53 7 -9
58 -124 19 58 3 -10
58 124 -19 58 7 -4
58 44 61 58 3 4
58 -44 -61 58 7 -10
61 109 41 61 5 1
61 -109 -41 61 6 -11
61 -131 -19 61 5 -11
61 131 19 61 6 -1
65 -31 71 65 1 7
65 31 -71 65 8 -9
65 79 61 65 4 3
65 -79 -61 65 7 -11
73 -151 31 73 3 -11
73 151 -31 73 8 -5
73 41 79 73 3 5
73 -41 -79 73 8 -11
74 116 59 74 5 2
74 -116 -59 74 7 -12
74 -164 -11 74 5 -12
74 164 11 74 7 -2
82 -116 71 82 1 -10
82 116 -71 82 9 -8
82 -44 89 82 1 8
82 44 -89 82 9 -10
85 -149 59 85 2 -11
85 149 -59 85 9 -7
85 -181 -29 85 6 -13
85 181 29 85 7 -1
89 121 79 89 5 3
89 -121 -79 89 8 -13
89 -199 -1 89 5 -13
89 199 1 89 8 -3
97 -209 29 97 4 -13
97 209 -29 97 9 -5
97 79 101 97 4 5
97 -79 -101 97 9 -13
101 139 -89 101 10 -9
101 -139 89 101 1 -11
101 59 -109 101 10 -11
101 -59 109 101 1 9
106 124 101 106 5 4
106 -124 -101 106 9 -14
106 -236 11 106 5 -14
106 236 -11 106 9 -4
109 211 -61 109 10 -7
109 -211 61 109 3 -13
109 -29 -121 109 10 -13
109 29 121 109 3 7
113 209 71 113 7 1
113 -209 -71 113 8 -15
113 -239 -41 113 7 -15
113 239 41 113 8 -1
122 164 -109 122 11 -10
122 -164 109 122 1 -12
122 -76 131 122 1 10
122 76 -131 122 11 -12
125 29 -139 125 11 -13
125 -29 139 125 2 9
130 244 -79 130 11 -8
130 -244 79 130 3 -14
130 -284 -31 130 7 -16
130 284 31 130 9 -2
137 281 -61 137 11 -7
137 -281 61 137 4 -15
137 -71 -149 137 11 -15
137 71 149 137 4 7
145 -191 131 145 1 -13
145 191 -131 145 12 -11
145 271 89 145 8 1
145 -271 -89 145 9 -17
146 -124 -151 146 11 -16
146 124 151 146 5 6
146 316 -41 146 11 -6
146 -316 41 146 5 -16
149 -229 -121 149 10 -17
149 229 121 149 7 3
149 331 19 149 10 -3
149 -331 -19 149 7 -17
|z| x y z u v
=================================================
answered Jul 21 at 17:41
Will Jagy
97.1k594196
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
add a comment |Â
up vote
1
down vote
The methods of Fricke and Klein (1897) show that a complete parametrization is given by a finite number of this type:
$$ x = 4 u^2 + 2 uv - v^2 ; , ; ; y = u^2 + 3 uv + v^2 ; , ; ; z = 2u^2 + 2 uv + v^2 ; , ; ; $$
where we do not restrict $u,v$ to positive, then freely use absolute values for $x,y,z.$ I will need to check whether this one recipe gives them all. If you want $x,y,z$ to have a common factor, simply multiply them all by some number.
Oh, note that $z$ comes out positive regardless, and we can bound $u,v,$ if we are bounding $z, ;$ since $2u^2 + 2uv+v^2 = u^2 + (u+v)^2 ; .$
Note that stereographic projection around a single point easily gives all rational solutions. However, there is then the problem of finding the primitive integer solutions.
One is enough. The above is sufficient. Oh, I was forgetting, might as well take $u geq 0$ to reduce repetition.
Sat Jul 21 11:06:37 PDT 2018
|z| x y z u v
1 -1 1 1 0 1
1 -1 1 1 0 -1
1 1 -1 1 1 -1
2 4 1 2 1 0
2 -4 -1 2 1 -2
5 -11 1 5 1 -3
5 11 -1 5 2 -1
10 4 11 10 1 2
10 -4 -11 10 3 -4
13 19 11 13 2 1
13 -19 -11 13 3 -5
13 -29 -1 13 2 -5
13 29 1 13 3 -1
17 1 19 17 1 3
17 -1 -19 17 4 -5
17 -31 11 17 1 -5
17 31 -11 17 4 -3
25 41 19 25 3 1
25 -41 -19 25 4 -7
26 -4 29 26 1 4
26 4 -29 26 5 -6
26 -44 19 26 1 -6
26 44 -19 26 5 -4
29 19 31 29 2 3
29 -19 -31 29 5 -7
29 -61 11 29 2 -7
29 61 -11 29 5 -3
34 44 31 34 3 2
34 -44 -31 34 5 -8
34 -76 1 34 3 -8
34 76 -1 34 5 -2
37 -11 41 37 1 5
37 11 -41 37 6 -7
37 -59 29 37 1 -7
37 59 -29 37 6 -5
41 71 29 41 4 1
41 -71 -29 41 5 -9
41 -89 -11 41 4 -9
41 89 11 41 5 -1
50 -76 41 50 1 -8
50 76 -41 50 7 -6
53 -101 31 53 2 -9
53 101 -31 53 7 -5
53 11 59 53 2 5
53 -11 -59 53 7 -9
58 -124 19 58 3 -10
58 124 -19 58 7 -4
58 44 61 58 3 4
58 -44 -61 58 7 -10
61 109 41 61 5 1
61 -109 -41 61 6 -11
61 -131 -19 61 5 -11
61 131 19 61 6 -1
65 -31 71 65 1 7
65 31 -71 65 8 -9
65 79 61 65 4 3
65 -79 -61 65 7 -11
73 -151 31 73 3 -11
73 151 -31 73 8 -5
73 41 79 73 3 5
73 -41 -79 73 8 -11
74 116 59 74 5 2
74 -116 -59 74 7 -12
74 -164 -11 74 5 -12
74 164 11 74 7 -2
82 -116 71 82 1 -10
82 116 -71 82 9 -8
82 -44 89 82 1 8
82 44 -89 82 9 -10
85 -149 59 85 2 -11
85 149 -59 85 9 -7
85 -181 -29 85 6 -13
85 181 29 85 7 -1
89 121 79 89 5 3
89 -121 -79 89 8 -13
89 -199 -1 89 5 -13
89 199 1 89 8 -3
97 -209 29 97 4 -13
97 209 -29 97 9 -5
97 79 101 97 4 5
97 -79 -101 97 9 -13
101 139 -89 101 10 -9
101 -139 89 101 1 -11
101 59 -109 101 10 -11
101 -59 109 101 1 9
106 124 101 106 5 4
106 -124 -101 106 9 -14
106 -236 11 106 5 -14
106 236 -11 106 9 -4
109 211 -61 109 10 -7
109 -211 61 109 3 -13
109 -29 -121 109 10 -13
109 29 121 109 3 7
113 209 71 113 7 1
113 -209 -71 113 8 -15
113 -239 -41 113 7 -15
113 239 41 113 8 -1
122 164 -109 122 11 -10
122 -164 109 122 1 -12
122 -76 131 122 1 10
122 76 -131 122 11 -12
125 29 -139 125 11 -13
125 -29 139 125 2 9
130 244 -79 130 11 -8
130 -244 79 130 3 -14
130 -284 -31 130 7 -16
130 284 31 130 9 -2
137 281 -61 137 11 -7
137 -281 61 137 4 -15
137 -71 -149 137 11 -15
137 71 149 137 4 7
145 -191 131 145 1 -13
145 191 -131 145 12 -11
145 271 89 145 8 1
145 -271 -89 145 9 -17
146 -124 -151 146 11 -16
146 124 151 146 5 6
146 316 -41 146 11 -6
146 -316 41 146 5 -16
149 -229 -121 149 10 -17
149 229 121 149 7 3
149 331 19 149 10 -3
149 -331 -19 149 7 -17
|z| x y z u v
=================================================
answered Jul 21 at 17:41
Will Jagy
97.1k594196
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The methods of Fricke and Klein (1897) show that a complete parametrization is given by a finite number of this type:
$$ x = 4 u^2 + 2 uv - v^2 ; , ; ; y = u^2 + 3 uv + v^2 ; , ; ; z = 2u^2 + 2 uv + v^2 ; , ; ; $$
where we do not restrict $u,v$ to positive, then freely use absolute values for $x,y,z.$ I will need to check whether this one recipe gives them all. If you want $x,y,z$ to have a common factor, simply multiply them all by some number.
Oh, note that $z$ comes out positive regardless, and we can bound $u,v,$ if we are bounding $z, ;$ since $2u^2 + 2uv+v^2 = u^2 + (u+v)^2 ; .$
Note that stereographic projection around a single point easily gives all rational solutions. However, there is then the problem of finding the primitive integer solutions.
One is enough. The above is sufficient. Oh, I was forgetting, might as well take $u geq 0$ to reduce repetition.
Sat Jul 21 11:06:37 PDT 2018
|z| x y z u v
1 -1 1 1 0 1
1 -1 1 1 0 -1
1 1 -1 1 1 -1
2 4 1 2 1 0
2 -4 -1 2 1 -2
5 -11 1 5 1 -3
5 11 -1 5 2 -1
10 4 11 10 1 2
10 -4 -11 10 3 -4
13 19 11 13 2 1
13 -19 -11 13 3 -5
13 -29 -1 13 2 -5
13 29 1 13 3 -1
17 1 19 17 1 3
17 -1 -19 17 4 -5
17 -31 11 17 1 -5
17 31 -11 17 4 -3
25 41 19 25 3 1
25 -41 -19 25 4 -7
26 -4 29 26 1 4
26 4 -29 26 5 -6
26 -44 19 26 1 -6
26 44 -19 26 5 -4
29 19 31 29 2 3
29 -19 -31 29 5 -7
29 -61 11 29 2 -7
29 61 -11 29 5 -3
34 44 31 34 3 2
34 -44 -31 34 5 -8
34 -76 1 34 3 -8
34 76 -1 34 5 -2
37 -11 41 37 1 5
37 11 -41 37 6 -7
37 -59 29 37 1 -7
37 59 -29 37 6 -5
41 71 29 41 4 1
41 -71 -29 41 5 -9
41 -89 -11 41 4 -9
41 89 11 41 5 -1
50 -76 41 50 1 -8
50 76 -41 50 7 -6
53 -101 31 53 2 -9
53 101 -31 53 7 -5
53 11 59 53 2 5
53 -11 -59 53 7 -9
58 -124 19 58 3 -10
58 124 -19 58 7 -4
58 44 61 58 3 4
58 -44 -61 58 7 -10
61 109 41 61 5 1
61 -109 -41 61 6 -11
61 -131 -19 61 5 -11
61 131 19 61 6 -1
65 -31 71 65 1 7
65 31 -71 65 8 -9
65 79 61 65 4 3
65 -79 -61 65 7 -11
73 -151 31 73 3 -11
73 151 -31 73 8 -5
73 41 79 73 3 5
73 -41 -79 73 8 -11
74 116 59 74 5 2
74 -116 -59 74 7 -12
74 -164 -11 74 5 -12
74 164 11 74 7 -2
82 -116 71 82 1 -10
82 116 -71 82 9 -8
82 -44 89 82 1 8
82 44 -89 82 9 -10
85 -149 59 85 2 -11
85 149 -59 85 9 -7
85 -181 -29 85 6 -13
85 181 29 85 7 -1
89 121 79 89 5 3
89 -121 -79 89 8 -13
89 -199 -1 89 5 -13
89 199 1 89 8 -3
97 -209 29 97 4 -13
97 209 -29 97 9 -5
97 79 101 97 4 5
97 -79 -101 97 9 -13
101 139 -89 101 10 -9
101 -139 89 101 1 -11
101 59 -109 101 10 -11
101 -59 109 101 1 9
106 124 101 106 5 4
106 -124 -101 106 9 -14
106 -236 11 106 5 -14
106 236 -11 106 9 -4
109 211 -61 109 10 -7
109 -211 61 109 3 -13
109 -29 -121 109 10 -13
109 29 121 109 3 7
113 209 71 113 7 1
113 -209 -71 113 8 -15
113 -239 -41 113 7 -15
113 239 41 113 8 -1
122 164 -109 122 11 -10
122 -164 109 122 1 -12
122 -76 131 122 1 10
122 76 -131 122 11 -12
125 29 -139 125 11 -13
125 -29 139 125 2 9
130 244 -79 130 11 -8
130 -244 79 130 3 -14
130 -284 -31 130 7 -16
130 284 31 130 9 -2
137 281 -61 137 11 -7
137 -281 61 137 4 -15
137 -71 -149 137 11 -15
137 71 149 137 4 7
145 -191 131 145 1 -13
145 191 -131 145 12 -11
145 271 89 145 8 1
145 -271 -89 145 9 -17
146 -124 -151 146 11 -16
146 124 151 146 5 6
146 316 -41 146 11 -6
146 -316 41 146 5 -16
149 -229 -121 149 10 -17
149 229 121 149 7 3
149 331 19 149 10 -3
149 -331 -19 149 7 -17
|z| x y z u v
=================================================
answered Jul 21 at 17:41
Will Jagy
97.1k594196
The methods of Fricke and Klein (1897) show that a complete parametrization is given by a finite number of this type:
$$ x = 4 u^2 + 2 uv - v^2 ; , ; ; y = u^2 + 3 uv + v^2 ; , ; ; z = 2u^2 + 2 uv + v^2 ; , ; ; $$
where we do not restrict $u,v$ to positive, then freely use absolute values for $x,y,z.$ I will need to check whether this one recipe gives them all. If you want $x,y,z$ to have a common factor, simply multiply them all by some number.
Oh, note that $z$ comes out positive regardless, and we can bound $u,v,$ if we are bounding $z, ;$ since $2u^2 + 2uv+v^2 = u^2 + (u+v)^2 ; .$
Note that stereographic projection around a single point easily gives all rational solutions. However, there is then the problem of finding the primitive integer solutions.
One is enough. The above is sufficient. Oh, I was forgetting, might as well take $u geq 0$ to reduce repetition.
Sat Jul 21 11:06:37 PDT 2018
|z| x y z u v
1 -1 1 1 0 1
1 -1 1 1 0 -1
1 1 -1 1 1 -1
2 4 1 2 1 0
2 -4 -1 2 1 -2
5 -11 1 5 1 -3
5 11 -1 5 2 -1
10 4 11 10 1 2
10 -4 -11 10 3 -4
13 19 11 13 2 1
13 -19 -11 13 3 -5
13 -29 -1 13 2 -5
13 29 1 13 3 -1
17 1 19 17 1 3
17 -1 -19 17 4 -5
17 -31 11 17 1 -5
17 31 -11 17 4 -3
25 41 19 25 3 1
25 -41 -19 25 4 -7
26 -4 29 26 1 4
26 4 -29 26 5 -6
26 -44 19 26 1 -6
26 44 -19 26 5 -4
29 19 31 29 2 3
29 -19 -31 29 5 -7
29 -61 11 29 2 -7
29 61 -11 29 5 -3
34 44 31 34 3 2
34 -44 -31 34 5 -8
34 -76 1 34 3 -8
34 76 -1 34 5 -2
37 -11 41 37 1 5
37 11 -41 37 6 -7
37 -59 29 37 1 -7
37 59 -29 37 6 -5
41 71 29 41 4 1
41 -71 -29 41 5 -9
41 -89 -11 41 4 -9
41 89 11 41 5 -1
50 -76 41 50 1 -8
50 76 -41 50 7 -6
53 -101 31 53 2 -9
53 101 -31 53 7 -5
53 11 59 53 2 5
53 -11 -59 53 7 -9
58 -124 19 58 3 -10
58 124 -19 58 7 -4
58 44 61 58 3 4
58 -44 -61 58 7 -10
61 109 41 61 5 1
61 -109 -41 61 6 -11
61 -131 -19 61 5 -11
61 131 19 61 6 -1
65 -31 71 65 1 7
65 31 -71 65 8 -9
65 79 61 65 4 3
65 -79 -61 65 7 -11
73 -151 31 73 3 -11
73 151 -31 73 8 -5
73 41 79 73 3 5
73 -41 -79 73 8 -11
74 116 59 74 5 2
74 -116 -59 74 7 -12
74 -164 -11 74 5 -12
74 164 11 74 7 -2
82 -116 71 82 1 -10
82 116 -71 82 9 -8
82 -44 89 82 1 8
82 44 -89 82 9 -10
85 -149 59 85 2 -11
85 149 -59 85 9 -7
85 -181 -29 85 6 -13
85 181 29 85 7 -1
89 121 79 89 5 3
89 -121 -79 89 8 -13
89 -199 -1 89 5 -13
89 199 1 89 8 -3
97 -209 29 97 4 -13
97 209 -29 97 9 -5
97 79 101 97 4 5
97 -79 -101 97 9 -13
101 139 -89 101 10 -9
101 -139 89 101 1 -11
101 59 -109 101 10 -11
101 -59 109 101 1 9
106 124 101 106 5 4
106 -124 -101 106 9 -14
106 -236 11 106 5 -14
106 236 -11 106 9 -4
109 211 -61 109 10 -7
109 -211 61 109 3 -13
109 -29 -121 109 10 -13
109 29 121 109 3 7
113 209 71 113 7 1
113 -209 -71 113 8 -15
113 -239 -41 113 7 -15
113 239 41 113 8 -1
122 164 -109 122 11 -10
122 -164 109 122 1 -12
122 -76 131 122 1 10
122 76 -131 122 11 -12
125 29 -139 125 11 -13
125 -29 139 125 2 9
130 244 -79 130 11 -8
130 -244 79 130 3 -14
130 -284 -31 130 7 -16
130 284 31 130 9 -2
137 281 -61 137 11 -7
137 -281 61 137 4 -15
137 -71 -149 137 11 -15
137 71 149 137 4 7
145 -191 131 145 1 -13
145 191 -131 145 12 -11
145 271 89 145 8 1
145 -271 -89 145 9 -17
146 -124 -151 146 11 -16
146 124 151 146 5 6
146 316 -41 146 11 -6
146 -316 41 146 5 -16
149 -229 -121 149 10 -17
149 229 121 149 7 3
149 331 19 149 10 -3
149 -331 -19 149 7 -17
|z| x y z u v
=================================================
answered Jul 21 at 17:41
Will Jagy
97.1k594196
edited Jul 21 at 18:15
answered Jul 21 at 17:41
Will Jagy
97.1k594196
answered Jul 21 at 17:41
Will Jagy
97.1k594196
answered Jul 21 at 17:41
Will Jagy
97.1k594196
97.1k594196
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
add a comment |Â
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
@HangWu there is a common factor of $19.$ Not primitive.
– Will Jagy
Jul 21 at 18:17
add a comment |Â
up vote
0
down vote
Do you know how to do this for Pythagorean triples? That is, do you know how to find solutions to $x^2 + y^2 = z^2$? This is equivalent to finding rational points on the circle $X^2 + Y^2 = 1$, and then taking $(x/z,y/z) = (X,Y)$. The way to find rational points on the circle is as follows: for a point on the circle $(X,Y)$, consider the line from $(-1,0)$ to $(X,Y)$. If this line has rational slope, then the equation for where the line meets the circle has rational coefficients and has the rational root at $(-1,0)$. The rational root test then tells you that the other root must be rational, i.e. $(X,Y)$ is rational.
Conversely, if $(X,Y)$ is rational, then the line from $(-1,0)$ to $(X,Y)$ has rational slope as well. Therefore, finding rational points on the unit circle are equivalent to looking at where the lines of rational slope at $(-1,0)$ intersect the circle; this can be done using the quadratic formula.
The same can be used for your equation. Simply consider all lines of rational slope starting at some fixed rational point of $X^2 / 5 + 4 Y^2 / 5 = 1$, e.g. $(1,1)$.
answered Jul 21 at 17:42
Marcus M
8,1731847
add a comment |Â
up vote
0
down vote
Do you know how to do this for Pythagorean triples? That is, do you know how to find solutions to $x^2 + y^2 = z^2$? This is equivalent to finding rational points on the circle $X^2 + Y^2 = 1$, and then taking $(x/z,y/z) = (X,Y)$. The way to find rational points on the circle is as follows: for a point on the circle $(X,Y)$, consider the line from $(-1,0)$ to $(X,Y)$. If this line has rational slope, then the equation for where the line meets the circle has rational coefficients and has the rational root at $(-1,0)$. The rational root test then tells you that the other root must be rational, i.e. $(X,Y)$ is rational.
Conversely, if $(X,Y)$ is rational, then the line from $(-1,0)$ to $(X,Y)$ has rational slope as well. Therefore, finding rational points on the unit circle are equivalent to looking at where the lines of rational slope at $(-1,0)$ intersect the circle; this can be done using the quadratic formula.
The same can be used for your equation. Simply consider all lines of rational slope starting at some fixed rational point of $X^2 / 5 + 4 Y^2 / 5 = 1$, e.g. $(1,1)$.
answered Jul 21 at 17:42
Marcus M
8,1731847
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Do you know how to do this for Pythagorean triples? That is, do you know how to find solutions to $x^2 + y^2 = z^2$? This is equivalent to finding rational points on the circle $X^2 + Y^2 = 1$, and then taking $(x/z,y/z) = (X,Y)$. The way to find rational points on the circle is as follows: for a point on the circle $(X,Y)$, consider the line from $(-1,0)$ to $(X,Y)$. If this line has rational slope, then the equation for where the line meets the circle has rational coefficients and has the rational root at $(-1,0)$. The rational root test then tells you that the other root must be rational, i.e. $(X,Y)$ is rational.
Conversely, if $(X,Y)$ is rational, then the line from $(-1,0)$ to $(X,Y)$ has rational slope as well. Therefore, finding rational points on the unit circle are equivalent to looking at where the lines of rational slope at $(-1,0)$ intersect the circle; this can be done using the quadratic formula.
The same can be used for your equation. Simply consider all lines of rational slope starting at some fixed rational point of $X^2 / 5 + 4 Y^2 / 5 = 1$, e.g. $(1,1)$.
answered Jul 21 at 17:42
Marcus M
8,1731847
Do you know how to do this for Pythagorean triples? That is, do you know how to find solutions to $x^2 + y^2 = z^2$? This is equivalent to finding rational points on the circle $X^2 + Y^2 = 1$, and then taking $(x/z,y/z) = (X,Y)$. The way to find rational points on the circle is as follows: for a point on the circle $(X,Y)$, consider the line from $(-1,0)$ to $(X,Y)$. If this line has rational slope, then the equation for where the line meets the circle has rational coefficients and has the rational root at $(-1,0)$. The rational root test then tells you that the other root must be rational, i.e. $(X,Y)$ is rational.
Conversely, if $(X,Y)$ is rational, then the line from $(-1,0)$ to $(X,Y)$ has rational slope as well. Therefore, finding rational points on the unit circle are equivalent to looking at where the lines of rational slope at $(-1,0)$ intersect the circle; this can be done using the quadratic formula.
The same can be used for your equation. Simply consider all lines of rational slope starting at some fixed rational point of $X^2 / 5 + 4 Y^2 / 5 = 1$, e.g. $(1,1)$.
answered Jul 21 at 17:42
Marcus M
8,1731847
answered Jul 21 at 17:42
Marcus M
8,1731847
answered Jul 21 at 17:42
Marcus M
8,1731847
answered Jul 21 at 17:42
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
1
math.stackexchange.com/questions/2773097/…
– individ
Jul 21 at 17:51