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

Question

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?

Answer 1

$u^3=(r+s)(r-s)$ 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 $$r-s=b^3$$

Answer 2

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.

Answer 3

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

Answer 4

We can find one or more Pythagorean triples for any odd leg $\ge 3$using a function of $(m,A)$: $$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$

This is useful in finding $x^3+B^2=C^2$ because we can plug in any odd cube $(A)$ an make a small finite search of values based on $A$. We demonstrate with the first odd cube.

$$m_{min}=\lceil\sqrt{27}\space\rceil=\lceil 5.196152423\rceil=6\qquad \qquad m_{max}=\frac{A+1}{2}=\frac{28}{2}=14$$

Testing for $6\le m\le 14$, we find integers for $(m,n)=(6,3)\text{ and }(14,13)$

$$\text{For }(6,3)\quad A=6^2-3^2=27\quad B=2*6*3=36\quad C=6^2+3^2=45\qquad (3,36,45)$$

$$\text{For }(14,13), A=14^2-13^2=27, B=2*14*13=364, C=14^2+13^2=365\quad (3,364,365)$$ Here are triples generated by the same algorithm(s) for other cubes $$\text{For }5^3=125\qquad f(15,10)=(5,300,325)$$ $$\text{For }7^3=343\qquad f(28,21)=(7,1176,1225$$

$$\text{For }9^3=729\quad f(45,36)=(9,3240,3321)\qquad f(123,120)=(9,29520,29529)$$

$$\text{For }11^3=1331\qquad f(66,55)=(11,7260,7381)$$