Mixing
Solving a cubic diophantine equation $ax^3+bx^2=cy^3$
$begingroup$
I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.
Edit: How would the answer change if a and c are both 1, while b is a positive integer?
diophantine-equations cubic-equations
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
add a comment |
$begingroup$
I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.
Edit: How would the answer change if a and c are both 1, while b is a positive integer?
diophantine-equations cubic-equations
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
add a comment |
$begingroup$
I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.
Edit: How would the answer change if a and c are both 1, while b is a positive integer?
diophantine-equations cubic-equations
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
$endgroup$
I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.
Edit: How would the answer change if a and c are both 1, while b is a positive integer?
diophantine-equations cubic-equations
diophantine-equations cubic-equations
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
edited Jan 17 at 22:56
LittleKnownMathematician
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
asked Jan 17 at 13:07
LittleKnownMathematicianLittleKnownMathematician
507
507
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
frac{1}{p}+frac{1}{q}+frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:
F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:
Integer solutions of $x^3+y^3=z^2$
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
$endgroup$
add a comment |
$begingroup$
$$ x^3 + y^3 = z^2 $$
1 pic.
$$ y = a x $$
$$ (3 a - 2 b) x = b^2 - 3 a^2 $$
answered Jan 17 at 18:41
S. I.S. I.
112
$endgroup$
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
add a comment |
$begingroup$
This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: xi: 0]$, with $xi^3=(c/a)$, which are clearly 3.
So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.
I expect there is no general formula for this solutions.
Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).
The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.
Edit
For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.
Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.
I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.
answered Jan 17 at 21:47
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
Equation, $ax^3+bx^2=cy^3$
"SI" above has given parametric solution (and is shown below):
$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$
For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:
$(a,b,c)=(1,35,1)$
$(x,y)=(5,10)$
Another numerical solution is:
$2(6)^3+4(6)^2=9(4)^3$
answered Jan 19 at 4:14
SamSam
1
$endgroup$
$begingroup$
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
frac{1}{p}+frac{1}{q}+frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:
F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:
Integer solutions of $x^3+y^3=z^2$
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
$endgroup$
add a comment |
$begingroup$
This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
frac{1}{p}+frac{1}{q}+frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:
F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:
Integer solutions of $x^3+y^3=z^2$
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
$endgroup$
add a comment |
$begingroup$
This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
frac{1}{p}+frac{1}{q}+frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:
F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:
Integer solutions of $x^3+y^3=z^2$
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
$endgroup$
This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
frac{1}{p}+frac{1}{q}+frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:
F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.
Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:
Integer solutions of $x^3+y^3=z^2$
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
edited Jan 17 at 15:44
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
answered Jan 17 at 13:25
Dietrich BurdeDietrich Burde
79.7k647103
79.7k647103
add a comment |
add a comment |
$begingroup$
$$ x^3 + y^3 = z^2 $$
1 pic.
$$ y = a x $$
$$ (3 a - 2 b) x = b^2 - 3 a^2 $$
answered Jan 17 at 18:41
S. I.S. I.
112
$endgroup$
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
add a comment |
$begingroup$
$$ x^3 + y^3 = z^2 $$
1 pic.
$$ y = a x $$
$$ (3 a - 2 b) x = b^2 - 3 a^2 $$
answered Jan 17 at 18:41
S. I.S. I.
112
$endgroup$
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
add a comment |
$begingroup$
$$ x^3 + y^3 = z^2 $$
1 pic.
$$ y = a x $$
$$ (3 a - 2 b) x = b^2 - 3 a^2 $$
answered Jan 17 at 18:41
S. I.S. I.
112
$endgroup$
$$ x^3 + y^3 = z^2 $$
1 pic.
$$ y = a x $$
$$ (3 a - 2 b) x = b^2 - 3 a^2 $$
answered Jan 17 at 18:41
S. I.S. I.
112
answered Jan 17 at 18:41
S. I.S. I.
112
answered Jan 17 at 18:41
S. I.S. I.
112
answered Jan 17 at 18:41
S. I.S. I.
112
112
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
add a comment |
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
Note that none of them satisfies $z=x$, since we need $x^3+y^3=x^2$, and not $z^2$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:14
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$y=ax$ is not the right transformation. We need $z=ax$.
$endgroup$
– Dietrich Burde
Jan 17 at 19:34
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
$begingroup$
$$ a (s n k^3)^3 + (n (e h^3 - a s k^3)) (s n k^3)^2 = (s s e) (n k^2 h)^3 $$
$endgroup$
– S. I.
Jan 18 at 18:29
add a comment |
$begingroup$
This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: xi: 0]$, with $xi^3=(c/a)$, which are clearly 3.
So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.
I expect there is no general formula for this solutions.
Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).
The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.
Edit
For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.
Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.
I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.
answered Jan 17 at 21:47
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: xi: 0]$, with $xi^3=(c/a)$, which are clearly 3.
So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.
I expect there is no general formula for this solutions.
Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).
The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.
Edit
For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.
Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.
I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.
answered Jan 17 at 21:47
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: xi: 0]$, with $xi^3=(c/a)$, which are clearly 3.
So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.
I expect there is no general formula for this solutions.
Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).
The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.
Edit
For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.
Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.
I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.
answered Jan 17 at 21:47
xarlesxarles
1,50079
$endgroup$
This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: xi: 0]$, with $xi^3=(c/a)$, which are clearly 3.
So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.
I expect there is no general formula for this solutions.
Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).
The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.
Edit
For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.
Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.
I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.
answered Jan 17 at 21:47
xarlesxarles
1,50079
edited Jan 18 at 10:52
answered Jan 17 at 21:47
xarlesxarles
1,50079
answered Jan 17 at 21:47
xarlesxarles
1,50079
answered Jan 17 at 21:47
xarlesxarles
1,50079
1,50079
add a comment |
add a comment |
$begingroup$
Equation, $ax^3+bx^2=cy^3$
"SI" above has given parametric solution (and is shown below):
$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$
For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:
$(a,b,c)=(1,35,1)$
$(x,y)=(5,10)$
Another numerical solution is:
$2(6)^3+4(6)^2=9(4)^3$
answered Jan 19 at 4:14
SamSam
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
add a comment |
$begingroup$
Equation, $ax^3+bx^2=cy^3$
"SI" above has given parametric solution (and is shown below):
$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$
For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:
$(a,b,c)=(1,35,1)$
$(x,y)=(5,10)$
Another numerical solution is:
$2(6)^3+4(6)^2=9(4)^3$
answered Jan 19 at 4:14
SamSam
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
add a comment |
$begingroup$
Equation, $ax^3+bx^2=cy^3$
"SI" above has given parametric solution (and is shown below):
$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$
For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:
$(a,b,c)=(1,35,1)$
$(x,y)=(5,10)$
Another numerical solution is:
$2(6)^3+4(6)^2=9(4)^3$
answered Jan 19 at 4:14
SamSam
1
$endgroup$
Equation, $ax^3+bx^2=cy^3$
"SI" above has given parametric solution (and is shown below):
$a (s n k^3)^3 + (ne h^3 - an s k^3) (s n k^3)^2 = (s^2e) (n k^2 h)^3$
For $(a,s,n,k,e,h)=(3,1,5,1,3,2)$ the solution is:
$(a,b,c)=(1,35,1)$
$(x,y)=(5,10)$
Another numerical solution is:
$2(6)^3+4(6)^2=9(4)^3$
answered Jan 19 at 4:14
SamSam
1
answered Jan 19 at 4:14
SamSam
1
answered Jan 19 at 4:14
SamSam
1
answered Jan 19 at 4:14
SamSam
1
1
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 4:23
add a comment |
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