Variation of Pythagorean triplets: $x^2+y^2 = z^3$

I need to prove that the equation $x^2 + y^2 = z^3$ has infinitely many solutions for positive $x, y$ and $z$. I got to as far as $4^3 = 8^2$ but that seems to be of no help.

Can some one help me with it?

Solution 1:

Take any Pythagorean triplet $(a,b,c)$.

$$\begin{align*} a^2+b^2 &=c^2\\ a^2\cdot c^4+b^2\cdot c^4&=(c^2)^3\\ (ac^2)^2+(bc^2)^2 &=(c^2)^3 \end{align*}$$

Multiplying $c^{6k-2}$, where $k$ is a natural number.

Solution 2:

As an alternative, you can take any standard Pythagorean triple, e.g. $3^2+4^2=5^2$, and then multiply through by $5^4$ to get:

$$3^2.5^4 + 4^2.5^4 = 5^6$$

i.e.

$$(3.5^2)^2 + (4.5^2)^2 = (5^2)^3$$

which will give an infinite set of solutions.

Solution 3:

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

Solution 4:

Set $z=a^2+b^2=(a+bi)(a-bi)$, $(i=\sqrt{-1})$

$x^2+y^2=(x+yi)(x-yi)=z^3=(a+bi)^3(a-bi)^3,$

$(1)\quad x+yi=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i,$

$x=a^3-3ab^2,y=3a^2b-b^3,z=a^2+b^2.$

$(2)\quad x+yi=(a+bi)^2(a-bi)=(a^2+b^2)(a+bi),$

$x=(a^2+b^2)a,y=(a^2+b^2)b,z=a^2+b^2$

Solution 5:

There are already infinitely many solutions among $2^k$'s: $$2^k+2^k=2^{k+1}$$ So, if $k$ is even and $3\,|\,k+1$ (that is, $k\equiv 2\pmod{6}$), then it's a solution.