Diophantine equation $x^2 + y^2 = z^3$

Question

I have found all solutions to the Diophantine equation $x^2 + y^2 = z^3$ when $z$ is odd. I am having some difficulty finding the solutions when $z$ is even. I am asking for a proof that provides the solutions where $z$ is even. I want the proof to be elementary and use only Number theory and perhaps Calculus or basic ideas about groups and rings.

Answer 1

Unfortunately, there isn't (apparently) one complete polynomial parameterization to

$$x^2+y^2 = z^k\tag1$$

when $k>2$. For $k=2$, the complete solution is,

$$x,\,y,\,z = (a^2-b^2)s,\; (2ab)s,\; (a^2+b^2)s$$

where $s$ is a scaling factor. Using complex numbers $a+b i$, one can generalize the method. For $k=3$, it is

$$x,\,y,\,z = (a^3 - 3a b^2)s^3,\; (3a^2 b - b^3)s^3,\; (a^2+b^2)s^2\tag2$$

but you can no longer find rational $a,b,s$ for certain solutions. For example,

$\hskip2.7in$ $9^2+46^2 = 13^3\quad$ Yes

$\hskip2.7in$ $58^2+145^2=29^3\quad$ No

(You can click on the Yes/No links for Walpha output.) A related discussion can be found in this post while an alternative method is described here. For the case $k=3$, if $a^2+b^2=c^3$, then an infinite more can be found as,

$$(a u^3 + 3 b u^2 v - 3 a u v^2 - b v^3)^2 + (b u^3 - 3 a u^2 v - 3 b u v^2 + a v^3)^2 = c^3(u^2+v^2)^3\tag3$$

which should provide some solutions not covered by $(2)$.

Answer 2

Fermat's two squares theorem says exactly which integers $n$ can be written as the sum of two squares, and indeed it can be made constructive, with a procedure to find all such representations. I recommend applying that known procedure to $n=z^3$. I don't think there's a significantly easier way; for example, already when $z$ is a high power of $5$ (or twice a high power of $5$), there are many representations.

Answer 3

Playing with the degrees and undetermined coefficients, we try to solve $$r^2(\alpha r^2+\beta s^2)^2+s^2(\gamma\space r^2+\delta s^2)^2=(\lambda r^2+\mu s^2)^3$$ in order to get an identity like for the pythagorean triples.

Operating, $$\alpha^2 r^6+(2\alpha \beta+\gamma^2)r^4s^2+(\beta^2+2\gamma \delta)r^2s^4+\delta^2 s^6=(\lambda r^2+\mu s^2)^3$$

We see convenient at first sight take $\alpha^2=\delta^2=1$ so that $$\pm2\beta+\gamma^2=\beta^2\pm2\gamma=3$$ Finally we take the values $$(\alpha,\beta,\gamma,\delta,\lambda,\mu)=(1, -3, 3,-1, 1, 1)$$ an we have get the identity $$ [r(r^2-3s^2)]^2+[s(3r^2-s^2)]^2=(r^2+s^2)^3$$ From which infinitely many solutions.

Answer 4

from equation: $\left( {p}^{2}+{k}^{2}\right) \,{z}^{2}={y}^{2}+{x}^{2}$

$(-2\,a\,b\,p-{b}^{2}\,k+{a}^{2}\,k,\left( {a}^{2}-{b}^{2}\right) \,p+2\,a\,b\,k,{b}^{2}+{a}^{2})$ $(2\,a\,b\,p-{b}^{2}\,k+{a}^{2}\,k,\left( {a}^{2}-{b}^{2}\right) \,p-2\,a\,b\,k,{b}^{2}+{a}^{2})$

if $z={p}^{2}+{k}^{2}$

${\left( {p}^{3}+{k}^{2}\,p\right) }^{2}+{\left( k\,{p}^{2}+{k}^{3}\right) }^{2}={\left( {p}^{2}+{k}^{2}\right) }^{3}$

${\left( {p}^{3}-3\,{k}^{2}\,p\right) }^{2}+{\left( 3\,k\,{p}^{2}-{k}^{3}\right) }^{2}={\left( {p}^{2}+{k}^{2}\right) }^{3}$

Answer 5

Usually a view. http://www.artofproblemsolving.com/community/c3046h1124891_almost_pythagoras http://www.artofproblemsolving.com/community/c3046h1049554__

You can write the solution in this form. http://www.artofproblemsolving.com/community/c3046h1054060_cubes_with_squares

http://www.artofproblemsolving.com/community/c3046h1048815___ http://www.artofproblemsolving.com/community/c3046h1047876____

But usually use the standard simple approach. In the equation.

$$X^2+Y^2=Z^3$$

$$X=ab+cd$$

$$Y=cb-ad$$

And receive such record.

$$(a^2+c^2)(b^2+d^2)=Z*Z^2$$

$$b^2+d^2=Z^2$$

$$Z=a^2+c^2$$

So.

$$d=a^2-c^2$$

$$b=2ac$$

Then the decision on the record.

$$X=3ca^2-c^3$$

$$Y=3ac^2-a^3$$

$$Z=a^2+c^2$$