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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?

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

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  • $\begingroup$ 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
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$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$
...
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: $(\frac{n(n-1)}{2}, \frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...

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  • $\begingroup$ Oh, this is nice and brilliant! $\endgroup$ – Maged Saeed Dec 19 '18 at 15:50
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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 $x\equiv y \pmod 2$, 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 $x\equiv y \pmod 2$, 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 $x\equiv y \pmod 2$. 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.

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  • $\begingroup$ 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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