How to solve this Diophantine equation?
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
add a comment |
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
add a comment |
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...
calculus diophantine-equations natural-numbers
calculus diophantine-equations natural-numbers
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
edited Dec 29 '18 at 12:45
Martín Vacas Vignolo
3,790623
3,790623
asked Dec 29 '18 at 12:43
Yan Dashkow
241
asked Dec 29 '18 at 12:43
Yan Dashkow
241
asked Dec 29 '18 at 12:43
Yan Dashkow
241
241
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
add a comment |
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered Dec 30 '18 at 8:19
Sam
1
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered Dec 30 '18 at 18:09
Sam
1
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
add a comment |
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
add a comment |
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
This is a case of the generalized Fermat equation
$$
x^p+y^q=z^r.
$$
For $(p,q,r)=(3,4,2)$ we have $frac{1}{p}+frac{1}{q}+frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:
F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Further Reference: The generalized Fermat equation.
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
edited Dec 29 '18 at 13:46
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
answered Dec 29 '18 at 13:14
Dietrich Burde
78k64386
78k64386
add a comment |
add a comment |
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
Here is one simple parameterization. We have,
$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$
given the Pell equation $x^2-3y^2 =1$.
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
edited Dec 30 '18 at 2:21
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
answered Dec 29 '18 at 14:03
Tito Piezas III
26.9k365169
26.9k365169
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
@AlexD $r=2$ in this question.
– Dietrich Burde
Dec 29 '18 at 23:34
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$?
– Alex D
Dec 30 '18 at 11:21
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
@AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$.
– Dietrich Burde
Dec 30 '18 at 11:47
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)?
– Yan Dashkow
Dec 30 '18 at 11:55
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
@YanDashkow We find them using one of the parametrizations, see Beukers article.
– Dietrich Burde
Dec 30 '18 at 13:31
|
show 1 more comment
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered Dec 30 '18 at 8:19
Sam
1
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered Dec 30 '18 at 8:19
Sam
1
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered Dec 30 '18 at 8:19
Sam
1
Above equation shown below has parameterization:
$x^3+y^4=z^2$
The below parameterization has no restriction such as the
Pell equation condition demonstrated by Tito Piezas.
$x=(p)^2(-q)^3$
$y=(p)(q)^2(k-1)$
$z=(p)^2(q)^4(2k-3)$
where, $p=(k-2)$ and $q=(k^2-2)$
For $k=3$ we get :
$(-343)^3+(98)^4=(7203)^2$
answered Dec 30 '18 at 8:19
Sam
1
answered Dec 30 '18 at 8:19
Sam
1
answered Dec 30 '18 at 8:19
Sam
1
answered Dec 30 '18 at 8:19
Sam
1
1
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)?
– Yan Dashkow
Dec 30 '18 at 10:55
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered Dec 30 '18 at 18:09
Sam
1
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered Dec 30 '18 at 18:09
Sam
1
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered Dec 30 '18 at 18:09
Sam
1
"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$
Solution is:
$x=3p^3(8k^2-40k+50)$
$y=p^2(20k^2-104k+135)$
$z=p^4(2k-5)^2(116k^2-540k+621)$
Where, $p=(4k^2-27)$
For $k=(13/5)$, we get after removing common factors:
$6^3+5^4=29^2$
answered Dec 30 '18 at 18:09
Sam
1
answered Dec 30 '18 at 18:09
Sam
1
answered Dec 30 '18 at 18:09
Sam
1
answered Dec 30 '18 at 18:09
Sam
1
1
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
Wait, why can't k be whole?
– Yan Dashkow
Dec 30 '18 at 19:30
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
"OP" enquired about integer coefficent's for the parametric
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs
to substitute $k=(m/n)$ in the parametrization & the resulting
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$ & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 31 '18 at 11:49
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 31 '18 at 11:49
Sam
1
answered Dec 31 '18 at 11:49
Sam
1
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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