# 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?

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Hint: $4^3 = 8^2$, and similarly $16^3 = 64^2$, $64^3 = 512^2$, $\ldots$, so if you have one solution, how can you multiply $x$, $y$ and $z$ by something to get new solutions? –  ferson2020 Mar 19 '13 at 15:43
nope it is supposed to be z^3 –  noddy Mar 19 '13 at 15:44
@ferson2020 thats correct but i would need to prove that 8^2, 64^2 ... can be expressed as a sum of 2 squares, which i think is not possible. Can you help me with that –  noddy Mar 19 '13 at 15:46
The OP's result also follows from the Brahmagupta-Fibonacci identity which says that the set of integers expressible as the sum of two squares is closed under multiplication. In particular, it is closed under positive powers. –  Byron Schmuland Mar 19 '13 at 16:19
@noddy Take a single solution ($2^2 + 2^2 = 2^3$) and multiply $x$ and $y$ by something ($8, 64, 512, \ldots$) and $z$ by something else ($4, 16, 64, \ldots$) to get a new solution. –  ferson2020 Mar 19 '13 at 17:42

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.

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QED - Quite easily done ;) –  Michael Viktor Starberg Mar 19 '13 at 21:39

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.

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+1. This simple trick is so effective in mixed-exponent Fermat-type equations. It's the reason why the Fermat-Catalan conjecture requires solutions to be coprime integers. –  Erick Wong Mar 19 '13 at 15:58

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

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

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

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

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

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

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

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+1 This is my favorite solution. –  Byron Schmuland Mar 19 '13 at 16:11
This is the solution I was about to submit. –  Lubin Mar 19 '13 at 17:25
Elegant solution, +1. –  octatoan Mar 20 '13 at 12:23

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.

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More generally, if $m$ and $n$ are coprime, the diophantine equation $x^m+y^m=z^n$ has infinitely many integral solutions. See this question. Your question is the specific case of $m=2$ and $n=3$.

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If $a^2+b^2=c^3$ is one solution, then for all integers $t$ we have

$$(at^3)^2+(bt^3)^2=(ct^2)^3$$

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the equation:

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

Has the solutions:

$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$

$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$

$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$

.................................................................................................................................................

$Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$

$+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$

$-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$

.................................................................................................................................................

$Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$

$q,t,k$ - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.

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