Integer Solutions of the Equation $u^3 = r^2-s^2$












2












$begingroup$


The question says the following:




Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.




The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $gcd(r,s) = 1$and $r+s equiv 1 pmod {2}$



In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.



I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.



Any ideas?






number-theory elementary-number-theory diophantine-equations pythagorean-triples







share|cite|improve this question









$endgroup$












  • $begingroup$
    Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
    $endgroup$
    – individ
    Dec 19 '18 at 15:06






  • 2




    $begingroup$
    One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:06










  • $begingroup$
    @individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:10






  • 1




    $begingroup$
    @MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:18






  • 1




    $begingroup$
    Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 15:19


















2












$begingroup$


The question says the following:




Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.




The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $gcd(r,s) = 1$and $r+s equiv 1 pmod {2}$



In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.



I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.



Any ideas?






number-theory elementary-number-theory diophantine-equations pythagorean-triples







share|cite|improve this question









$endgroup$












  • $begingroup$
    Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
    $endgroup$
    – individ
    Dec 19 '18 at 15:06






  • 2




    $begingroup$
    One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:06










  • $begingroup$
    @individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:10






  • 1




    $begingroup$
    @MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:18






  • 1




    $begingroup$
    Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 15:19
















2












2








2





$begingroup$


The question says the following:




Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.




The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $gcd(r,s) = 1$and $r+s equiv 1 pmod {2}$



In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.



I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.



Any ideas?






number-theory elementary-number-theory diophantine-equations pythagorean-triples







share|cite|improve this question









$endgroup$




The question says the following:




Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.




The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $gcd(r,s) = 1$and $r+s equiv 1 pmod {2}$



In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.



I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.



Any ideas?






number-theory elementary-number-theory diophantine-equations pythagorean-triples




number-theory elementary-number-theory diophantine-equations pythagorean-triples






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 15:02









Maged SaeedMaged Saeed

8921417




8921417












  • $begingroup$
    Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
    $endgroup$
    – individ
    Dec 19 '18 at 15:06






  • 2




    $begingroup$
    One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:06










  • $begingroup$
    @individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:10






  • 1




    $begingroup$
    @MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:18






  • 1




    $begingroup$
    Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 15:19




















  • $begingroup$
    Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
    $endgroup$
    – individ
    Dec 19 '18 at 15:06






  • 2




    $begingroup$
    One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:06










  • $begingroup$
    @individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:10






  • 1




    $begingroup$
    @MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
    $endgroup$
    – Frpzzd
    Dec 19 '18 at 15:18






  • 1




    $begingroup$
    Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
    $endgroup$
    – Jyrki Lahtonen
    Dec 19 '18 at 15:19


















$begingroup$
Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
$endgroup$
– individ
Dec 19 '18 at 15:06




$begingroup$
Any cube can be represented by a difference of squares. $$x^3=(y-z)(y+z)$$
$endgroup$
– individ
Dec 19 '18 at 15:06




2




2




$begingroup$
One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
$endgroup$
– Frpzzd
Dec 19 '18 at 15:06




$begingroup$
One general class of solutions is given by $r=frac{u^2+u}{2}$ and $s=frac{u^2-u}{2}$, but I am fairly sure this is not an exhaustive solution set.
$endgroup$
– Frpzzd
Dec 19 '18 at 15:06












$begingroup$
@individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:10




$begingroup$
@individ let $y = 5, z= 1$, then $4*6=24$ which is not a cube. My point is that when will $(y-z)(y+z)$ is a cube?
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:10




1




1




$begingroup$
@MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
$endgroup$
– Frpzzd
Dec 19 '18 at 15:18




$begingroup$
@MagedSaeed Write $u^3=(r-s)(r+s)$. Since $r+s$ and $r-s$ must have the same parity, and $u$ and $u^2$ must have the same parity, we may let $u=r-s$ and $u^2=r+s$. The same can be done for any two divisors of $u^3$ that have the same parity.
$endgroup$
– Frpzzd
Dec 19 '18 at 15:18




1




1




$begingroup$
Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 15:19






$begingroup$
Maged, I'm sure individ meant that if you can write down a factorization, any factorization will do, $u^3=ab$ such that $a$ and $b$ have the same parity, then you can solve for $y$ and $z$ from the system $a=y-z$, $b=y+z$. The choice $a=x$, $b=x^2$ gives you the solution Frpzzd provided.
$endgroup$
– Jyrki Lahtonen
Dec 19 '18 at 15:19












3 Answers
3






active

oldest

votes


















1












$begingroup$

$u^3=(r+s)(r-s)$ and $gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.



So let $r+s=a^3$, $r-s=b^3$. Then
$$r=frac{a^3+b^3}2$$
$$s=frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.



Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:23



















1












$begingroup$

$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$

...

The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

Which means the solutions are pairs of this form: $(frac{n(n-1)}{2}, frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$

...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, this is nice and brilliant!
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:50



















0












$begingroup$

I assume you are allowing $u,r,s$ to be negative.



Let us substitute $r-s=a$ so that your equation is equivalent to
$$u^3=a(a+2s)$$
thus, if $u^3$ can be written in the form $u^3=xy$ where $xequiv y pmod 2$, then we may let $a=x$ and $a+2s=y$, and solve an easy system of equations obtain values for $r$ and $s$.



Thus, if $u^3=xy$ and $xequiv y pmod 2$, then $r=frac{x+y}{2}$ and $s=frac{x-y}{2}$ is a possible solution.



Let's try and find the number of solutions $(r,s)$ given the value of $u^3$. Each solution $(r,s)$ can be put into one-to-one correspondence with a pair $(x,y)$ satisfying $u^3=xy$ and $xequiv y pmod 2$. If $u$ is even, there are $(v_2(u^3)-1)d_o(u^3)$ such pairs, and if $u$ is odd, there are $d_o(u^3)$ such pairs (where $v_2(u^3)$ is the 2-adic valuation of $u^3$ and $d_o(u^3)$ is the number of odd divisors of $u^3$), which can be easily proven by "dividing up" the factors of $2$ in $u^3$ between $x$ and $y$.



Thus, given $u^3$, there are $d_o(u^3)$ solutions if $u$ is odd and $(v_2(u^3)-1)d_o(u^3)$ solutions if $u$ is even.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but this did not give explicit formulas for $r$ and $s$.
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:26












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$u^3=(r+s)(r-s)$ and $gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.



So let $r+s=a^3$, $r-s=b^3$. Then
$$r=frac{a^3+b^3}2$$
$$s=frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.



Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:23
















1












$begingroup$

$u^3=(r+s)(r-s)$ and $gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.



So let $r+s=a^3$, $r-s=b^3$. Then
$$r=frac{a^3+b^3}2$$
$$s=frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.



Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:23














1












1








1





$begingroup$

$u^3=(r+s)(r-s)$ and $gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.



So let $r+s=a^3$, $r-s=b^3$. Then
$$r=frac{a^3+b^3}2$$
$$s=frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.



Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$






share|cite|improve this answer











$endgroup$



$u^3=(r+s)(r-s)$ and $gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.



So let $r+s=a^3$, $r-s=b^3$. Then
$$r=frac{a^3+b^3}2$$
$$s=frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.



Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and have different parity. Indeed,
$$r+s=a^3$$
which is odd and coprime with
$$r-s=b^3$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 15:42

























answered Dec 19 '18 at 15:20









ajotatxeajotatxe

54.1k24190




54.1k24190












  • $begingroup$
    That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:23


















  • $begingroup$
    That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:23
















$begingroup$
That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:23




$begingroup$
That is what I was looking for. I just have scratched this on a paper and immediately found it as an answer of yours. :)
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:23











1












$begingroup$

$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$

...

The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

Which means the solutions are pairs of this form: $(frac{n(n-1)}{2}, frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$

...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, this is nice and brilliant!
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:50
















1












$begingroup$

$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$

...

The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

Which means the solutions are pairs of this form: $(frac{n(n-1)}{2}, frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$

...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, this is nice and brilliant!
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:50














1












1








1





$begingroup$

$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$

...

The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

Which means the solutions are pairs of this form: $(frac{n(n-1)}{2}, frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$

...






share|cite|improve this answer











$endgroup$



$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$

...

The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

Which means the solutions are pairs of this form: $(frac{n(n-1)}{2}, frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$

...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 15:38

























answered Dec 19 '18 at 15:30









VasyaVasya

4,2671618




4,2671618












  • $begingroup$
    Oh, this is nice and brilliant!
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:50


















  • $begingroup$
    Oh, this is nice and brilliant!
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:50
















$begingroup$
Oh, this is nice and brilliant!
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:50




$begingroup$
Oh, this is nice and brilliant!
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:50











0












$begingroup$

I assume you are allowing $u,r,s$ to be negative.



Let us substitute $r-s=a$ so that your equation is equivalent to
$$u^3=a(a+2s)$$
thus, if $u^3$ can be written in the form $u^3=xy$ where $xequiv y pmod 2$, then we may let $a=x$ and $a+2s=y$, and solve an easy system of equations obtain values for $r$ and $s$.



Thus, if $u^3=xy$ and $xequiv y pmod 2$, then $r=frac{x+y}{2}$ and $s=frac{x-y}{2}$ is a possible solution.



Let's try and find the number of solutions $(r,s)$ given the value of $u^3$. Each solution $(r,s)$ can be put into one-to-one correspondence with a pair $(x,y)$ satisfying $u^3=xy$ and $xequiv y pmod 2$. If $u$ is even, there are $(v_2(u^3)-1)d_o(u^3)$ such pairs, and if $u$ is odd, there are $d_o(u^3)$ such pairs (where $v_2(u^3)$ is the 2-adic valuation of $u^3$ and $d_o(u^3)$ is the number of odd divisors of $u^3$), which can be easily proven by "dividing up" the factors of $2$ in $u^3$ between $x$ and $y$.



Thus, given $u^3$, there are $d_o(u^3)$ solutions if $u$ is odd and $(v_2(u^3)-1)d_o(u^3)$ solutions if $u$ is even.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but this did not give explicit formulas for $r$ and $s$.
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:26
















0












$begingroup$

I assume you are allowing $u,r,s$ to be negative.



Let us substitute $r-s=a$ so that your equation is equivalent to
$$u^3=a(a+2s)$$
thus, if $u^3$ can be written in the form $u^3=xy$ where $xequiv y pmod 2$, then we may let $a=x$ and $a+2s=y$, and solve an easy system of equations obtain values for $r$ and $s$.



Thus, if $u^3=xy$ and $xequiv y pmod 2$, then $r=frac{x+y}{2}$ and $s=frac{x-y}{2}$ is a possible solution.



Let's try and find the number of solutions $(r,s)$ given the value of $u^3$. Each solution $(r,s)$ can be put into one-to-one correspondence with a pair $(x,y)$ satisfying $u^3=xy$ and $xequiv y pmod 2$. If $u$ is even, there are $(v_2(u^3)-1)d_o(u^3)$ such pairs, and if $u$ is odd, there are $d_o(u^3)$ such pairs (where $v_2(u^3)$ is the 2-adic valuation of $u^3$ and $d_o(u^3)$ is the number of odd divisors of $u^3$), which can be easily proven by "dividing up" the factors of $2$ in $u^3$ between $x$ and $y$.



Thus, given $u^3$, there are $d_o(u^3)$ solutions if $u$ is odd and $(v_2(u^3)-1)d_o(u^3)$ solutions if $u$ is even.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but this did not give explicit formulas for $r$ and $s$.
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:26














0












0








0





$begingroup$

I assume you are allowing $u,r,s$ to be negative.



Let us substitute $r-s=a$ so that your equation is equivalent to
$$u^3=a(a+2s)$$
thus, if $u^3$ can be written in the form $u^3=xy$ where $xequiv y pmod 2$, then we may let $a=x$ and $a+2s=y$, and solve an easy system of equations obtain values for $r$ and $s$.



Thus, if $u^3=xy$ and $xequiv y pmod 2$, then $r=frac{x+y}{2}$ and $s=frac{x-y}{2}$ is a possible solution.



Let's try and find the number of solutions $(r,s)$ given the value of $u^3$. Each solution $(r,s)$ can be put into one-to-one correspondence with a pair $(x,y)$ satisfying $u^3=xy$ and $xequiv y pmod 2$. If $u$ is even, there are $(v_2(u^3)-1)d_o(u^3)$ such pairs, and if $u$ is odd, there are $d_o(u^3)$ such pairs (where $v_2(u^3)$ is the 2-adic valuation of $u^3$ and $d_o(u^3)$ is the number of odd divisors of $u^3$), which can be easily proven by "dividing up" the factors of $2$ in $u^3$ between $x$ and $y$.



Thus, given $u^3$, there are $d_o(u^3)$ solutions if $u$ is odd and $(v_2(u^3)-1)d_o(u^3)$ solutions if $u$ is even.






share|cite|improve this answer









$endgroup$



I assume you are allowing $u,r,s$ to be negative.



Let us substitute $r-s=a$ so that your equation is equivalent to
$$u^3=a(a+2s)$$
thus, if $u^3$ can be written in the form $u^3=xy$ where $xequiv y pmod 2$, then we may let $a=x$ and $a+2s=y$, and solve an easy system of equations obtain values for $r$ and $s$.



Thus, if $u^3=xy$ and $xequiv y pmod 2$, then $r=frac{x+y}{2}$ and $s=frac{x-y}{2}$ is a possible solution.



Let's try and find the number of solutions $(r,s)$ given the value of $u^3$. Each solution $(r,s)$ can be put into one-to-one correspondence with a pair $(x,y)$ satisfying $u^3=xy$ and $xequiv y pmod 2$. If $u$ is even, there are $(v_2(u^3)-1)d_o(u^3)$ such pairs, and if $u$ is odd, there are $d_o(u^3)$ such pairs (where $v_2(u^3)$ is the 2-adic valuation of $u^3$ and $d_o(u^3)$ is the number of odd divisors of $u^3$), which can be easily proven by "dividing up" the factors of $2$ in $u^3$ between $x$ and $y$.



Thus, given $u^3$, there are $d_o(u^3)$ solutions if $u$ is odd and $(v_2(u^3)-1)d_o(u^3)$ solutions if $u$ is even.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 15:17









FrpzzdFrpzzd

23k841110




23k841110












  • $begingroup$
    Thanks, but this did not give explicit formulas for $r$ and $s$.
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:26


















  • $begingroup$
    Thanks, but this did not give explicit formulas for $r$ and $s$.
    $endgroup$
    – Maged Saeed
    Dec 19 '18 at 15:26
















$begingroup$
Thanks, but this did not give explicit formulas for $r$ and $s$.
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:26




$begingroup$
Thanks, but this did not give explicit formulas for $r$ and $s$.
$endgroup$
– Maged Saeed
Dec 19 '18 at 15:26


















draft saved

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