I'd like to ask how to obtain all integer solutions to the Diophantine equation $x^2+4y^2=5z^2$ using parameterization? Thanks.
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1https://math.stackexchange.com/questions/2773097/how-to-find-all-rational-solutions-of-x2-3y2-7/2788381#2788381 – individ Jul 21 '18 at 17:51
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Assume that you want the primitive solutions. I shall determine integer solutions up to sign changes.
Note that $x\equiv \pm y\pmod{5}$. We may assume that $x\equiv y\pmod{5}$ (otherwise, just flip the sign of $y$). Now, write $$z^2=\frac{x^2+4y^2}{5}=\left(\frac{x+4y}{5}\right)^2+\left(\frac{2x-2y}{5}\right)^2\,.$$ That is, using the knowledge from Pythagorian triples, we get that $$\frac{x+4y}{5}=m^2-n^2\,,\,\,\frac{2x-2y}{5}=2mn\,,\text{ and }z=m^2+n^2$$ for some coprime $m,n\in\mathbb{Z}$ 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).
Batominovski
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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 '18 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
=================================================
Will Jagy
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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. $(1,1)$.
Marcus M
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