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

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.

• I see $x = y = z = 2$ is a trivial solution to the system. – Jack Tiger Lam Dec 6 '15 at 0:50
• If $z$ is even, then $x^2 + y^2$ is divisible by $4.$ In turn, this means both $x,y$ are even, with $x \equiv y \pmod 4.$ With enough effort this should allow you to finish. – Will Jagy Dec 6 '15 at 1:02
• @WillJagy: $x\equiv y\pmod 4$? – Greg Martin Dec 6 '15 at 1:16
• If $(x,y,z)$ is a solution to $x^2+y^2=z^3$, then so is $$(8n^3x, 8n^3y, 4n^2z)$$ for any non-negative integer $n$. This relies on the fact that multiplying both sides of the equation by an even sixth power will yield another solution. This gives you an infinite number of even solutions for each solution that you find (you said you'd already found all the solutions with odd $z$). This probably doesn't yield all solutions with even $z$, but it may help. – Zubin Mukerjee Dec 6 '15 at 1:36
• @Zubin Mukerjee Let a,b be integers. Then $(a^3 - 3ab^2,3a^2b - b^3, a^2 + b^2) is a solution. These solutions contain all the odd solutions – Tanner Carawan Dec 6 '15 at 1:48 ## 5 Answers 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,$\hskip2.7in9^2+46^2 = 13^3\quad$Yes$\hskip2.7in58^2+145^2=29^3\quad$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$(2)$. • Is this a joke? $$58^2+145^2=29^3$$ $$(2*29)^2+(5*29)^2=29^3$$ $$2^2+5^2=29$$ So can any nonlinear equation is reduced to linear. – individ Dec 6 '15 at 8:05 • @individ: Sigh... It should not take you a few seconds to answer your own question. – Tito Piezas III Dec 6 '15 at 8:19 • @Tanner Carawan: I asked a more general version of your question. Kindly see this post. – Tito Piezas III Dec 6 '15 at 9:22 • @TitoPiezasIII: I looked at it. Thanks for answering my question. I have not yet studied enough Abstract Algebra to understand why there isn't a complete polynomial parametrization. – Tanner Carawan Dec 7 '15 at 23:00 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. You can write the solution in this form. http://www.artofproblemsolving.com/community/c3046h1054060_cubes_with_squares 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. $$X=3ca^2-c^3$$ $$Y=3ac^2-a^3$$ $$Z=a^2+c^2$$ • Is every solution of that form? Why can we say X = ab + cd and Y = cb - ad? Why must b^2 + d^2 = z^2 or z = a^2 + c^2? What is all this talk about "record"? – Tanner Carawan Dec 6 '15 at 7:52 • @TannerCarawan Because such a simple entry - allows you to simply solve an equation. So why not use it? – individ Dec 6 '15 at 8:08 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. • All this is good, but the equations do not solve. In the first formula, You actually set what type must have solutions. But the equation we haven't decided about this kind of know nothing. Then solve the equation - when we do not know the solution. – individ Dec 6 '15 at 7:51 • Do you want to explain me more, please. I don´t understand what you desire to tell me. And consider I have no pretended to have found all the solutions but infinitely many of them by an identity which It was not sure I could get (have in account, please, my English of Google translator is weak) – Piquito Dec 6 '15 at 13:07 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}\$

• You lead the same solutions as above. What's the point of this response? – individ Dec 6 '15 at 13:57