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Solve for integers $x, y$ and $z$:

$x^2 + y^2 = z^3.$

I tried manipulating by adding and subtracting $2xy$ , but it didn't give me any other information, except the fact that $z^3 - 2xy$ and $z^3+2xy$ are perfect squares.

This doesn't give us much information to work on. I don't know if my steps are correct, I do not know how to approach this problem.

Any help would be appreciated.

Servaes
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3 Answers3

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Let $C$, $D$, $S$ and $T$ be integers, and define \begin{eqnarray*} x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\ y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\ z&=&ab^2Z=(C^2+D^2)(S^2+T^2). \end{eqnarray*} Then a routine verification shows that $x^2+y^2=z^3$. I will show that every solution is of this form. Moreover, if we require $S$ and $T$ to be coprime and nonnegative, every solution will have precisely one such representation, making this a proper parametrization.

Let $x$, $y$ and $z$ be integers such that $$x^2+y^2=z^3.$$ First note that $x$ and $y$ are not both odd, as otherwise we get a contradiction by reducing mod $8$.

Let $d:=\gcd(x,y)$ and let $a$ and $b$ be integers such that $d=ab^3$ and $a$ is cube-free. Then $d^2=a^2b^6$ divides $z^3$ and hence $a$ divides $z$. Writing $x=au$, $y=av$ and $z=aw$ we see that $$a^3w^3=z^3=x^2+y^2=(ab^3u)^2+(ab^3v)^2=a^2b^6(u^2+v^2),$$ from which it follows that $b^2$ divides $w$ because $a$ is cube-free. So writing $x=ab^3X$, $y=ab^3Y$, $z=ab^2Z$ shows that $$X^2+Y^2=aZ^3,$$ where $X$ and $Y$ are coprime. Factoring in $\Bbb{Z}[i]$ then shows that $$aZ^3=(X+Yi)(X-Yi),$$ where $X$ and $Y$ are coprime and not both odd, so the two factors are coprime. Then $$X+Yi=(A+Bi)(U+Vi)^3,$$ for some integers $A$, $B$, $U$ and $V$ such that $\gcd(A,B)=\gcd(U,V)=1$ and $A^2+B^2=a$ and $U^2+V^2=Z$. Then \begin{eqnarray*} X&=&AU^3-3BU^2V-3AUV^2+BV^3,\\ Y&=&BU^3+3AU^2V-3BUV^2-AV^3, \end{eqnarray*} and hence for $C=bA$ and $D=bB$ we find that \begin{eqnarray*} x&=&ab^3X=(C^2+D^2)(CS^3-3DS^2T-3CST^2+DT^3),\\ y&=&ab^3Y=(C^2+D^2)(DS^3+3CS^2T-3DST^2-CT^3),\\ z&=&ab^2Z=(C^2+D^2)(S^2+T^2). \end{eqnarray*}

In particular, parametrizations given in the other answers and comments correspond to $(C,D,S,T)=$ $$(1,0,a,b),\qquad(k,k,1,0),\qquad(1,k,1,0),\qquad(a,b,1,0).$$

Servaes
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  • $2^2+2^2=2^3$, hence $(x,y,z)=(2,2,2)$ is a solution. Then $d=2$, $a=2$, $b=1$, $X=1$, $Y=1$, which is a contradiction to your claim that $X$ and $Y$ cannot be both odd. – Bogdan Grechuk Sep 07 '22 at 13:04
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Factoring over Gaussian integers $$(x+iy)(x-iy)=z^3$$ so it is sufficient (but not necessary) for $x+iy$ to be a cube. That is, $$\begin{align}x+iy&=(a+bi)^3\\ &=(a^3-3ab^2)+\left(3a^2b-b^3\right)i \end{align}$$

So choose any $a,b$ and set $x=a^3-3ab^2$, $y=3a^2b-b^3$. For example $a=5$, $b=7$ gives $x=-610$, $y=182$. And indeed, $(-610)^2+182^2=74^3$.

This recovers some solution families mentioned in the comments. For example with $a=-k, b=k$, it recovers $x=2k^3,y=2k^3,z=2k^2$.

However it does not produce all solutions. For example in the family $x=1+k^2,y=k+k^3$, there is the the solution $x=5,y=10$. But the family in this answer will not yield that as a solution.

2'5 9'2
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  • It is not necessary for $x+iy$ to be a cube, unless perhaps you require $x+iy$ and $x-iy$ to be coprime, i.e. that $\gcd(x,y)=1$ and $x\not\equiv y\pmod{2}$. – Servaes Apr 17 '21 at 15:08
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    And it's worth noting that then $z=a^2+b^2$. – Servaes Apr 17 '21 at 15:12
  • @Servaes That $z$ is the sum of two squares comes from the parity of the prime factors of the form $p=4n+3$ in $z^3$. One could in fact begin there with $z=a^2+b^2$ and $x^2+y^2=z^2(a^2+b^2)$. Then $z^2$ can be expressed (perhaps in many ways) as a sum of two squares, and these can be combined. – Mark Bennet Apr 17 '21 at 15:46
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    @MarkBennet That's true. What I was trying to get at, is that this answer gives a parametrization $$(x,y)=(a^3-3ab^2,3a^2b-b^3),$$ of some (though not all) solutions. It would be nice to complete this parametrization by also including $z$, giving $$(x,y,z)=(a^3-3ab^2,3a^2b-b^3,a^2+b^2).$$ – Servaes Apr 17 '21 at 16:43
  • As a final note, because $x$ and $y$ cannot both be odd, this only fails when $x$ and $y$ are not coprime. Such solutions correspond precisely to primitive solutions to $$u^2+v^2=dw^3,$$ where $d>1$ is the cube-free part of $\gcd(x,y)$. These are obtained in the same way, by factoring $d$ in $\Bbb{Z}[i]$. – Servaes Apr 17 '21 at 17:01
  • @Servaes The way I looked at it was to take $u^2+v^2=(dw)w^2$ and you need $dw$ as the sum of two squares. Solutions can be built fro appropriate choices of $d$ and $w$. – Mark Bennet Apr 18 '21 at 06:45
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First I search for trivial solutions: $(0,0,0)$ works. Likewise $(0,1,1)$ and $(1,0,1)$.

$z$ has to be non-negative, as its cube is a sum of squares. Likewise if a negative value of $x$ or $y$ worked, so would a positive value. So we can stick to positive $(x,y,z)$ since we already found the solutions with $0$'s.

So, next up would be $z=2$, so $x^2 + y^2 = 8$, which has a pretty immediate solution $(2,2,2)$.

Going to $z = 3$, we would need two squares that add up to $27$, and none work.

We could continue taking values of $z$ one at a time, or we can pick a constraint and see if we can find a set of solutions. Suppose $x = 1$. Are there any squares that are one less than a cube?

Comparing $1,4,9,16,25,36,49,64,81,100,121,144,169,196,225...$ with

$1,8,27,64,125,216,343,512,729,1000,1331,1728...$. I don't see any such option, but I do notice that $121 + 4 = 125$, so there is a solution $(2,11,5)$.

If we consider even vs odd possibilities: if $x$ and $y$ are both even, so must $z$ be. However we can't factor all the $2$'s out. If $(2a, 2b, 2c)$ is a solution, $(a, b, c)$ is not.

It looks like others are finding better patterns in the comments so I'll leave it there.

RobertTheTutor
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