# Integer solutions of $x^3+y^3=z^2$

Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?

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Can $x = y$? Because if so, $(2, 2, 4)$ is a solution too. –  Sim Apr 22 '13 at 23:55
I didn't see your comment. Note my answer below. I think we saw this at the same time. –  ncmathsadist Apr 23 '13 at 0:10
Have you tried a little computer experiment? –  ncmathsadist Apr 23 '13 at 0:11
$x=-y$ and $z=0$ would yield a bunch of solutions as well. –  JB King Apr 23 '13 at 22:13

Note that if $(x,y,z)$ is a solution, then so are $(y,x,z)$ and $(a^2x,a^2y,a^3z)$ where $a \in \mathbb{Z}$.

Below are some primitive solutions, you can use them and follow the above technique to generate infinite solutions. Hence, we will leave out these duplicates.

$$(1,2,3); (2,2,4); (2,46,312); (7,21,98); (10,65,525); (11,37,228); (14,70,588); (22,26,168)$$ and so on.

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Thanks everyone. Your responses have helped me found an approach. I will try positive integers f and g until f^3+g^3=(h^2)k. Then, a solution is obtained as (kf)^3+(kg)^3=(hk^2)^2. For example, from 3^3+5^3=4x38, I obtain 114^3+190^3=2888^2. As another example, I try 1^3+5^3=9x14. Then a solution is (14,70,588). –  Obinly Apr 24 '13 at 4:40

Much as for Pythagorean triples, there are parametrizations for the coprime solutions to such equations (this is generally true for $x^p+y^q=z^r$ whenever the sum of the reciprocals of $p, q$ and $r$ exceeds $1$). One can find these on pages 467 to 470 of Henri Cohen's excellent Springer GTM 240 : one such parametrization (there are $3$ in total) is to take $$x=-3s^4+6t^2s^2+t^4, \; y=3s^4+6t^2s^2-t^4, \; z = 6 s t ( 3 s^4+t^4),$$ where $s$ and $t$ are coprime, of opposite parity, and $3$ does not divide $t$. Of course, this is up to exchange of $x$ and $y$.

As noted earlier, non-coprime solutions are readily found.

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You can multiply $x$ and $y$ by $a^2$, and $z$ by $a^3$.

There is the boring $(0,0,0)$. And the almost equally boring $(-1, 1,0)$. And then we have $2^3+2^3=4^2$ and its relatives.

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Could you throw some light on how you found the above family? Also, did you mean $t^4$ (instead of $t * 4$) since the expressions for $x$ and $y$ are not symmetric in $t$ and $s$. –  user17762 Apr 23 '13 at 0:11
It is somewhat more complicated, there are conditions on $s$ and $t$. I copied too quickly from old files of mine. –  André Nicolas Apr 23 '13 at 0:31

I did a little experiment. I got these.

1 2 3
2 1 3
2 2 4
4 8 24
7 21 98
8 4 24
8 8 32
9 18 81
18 9 81
21 7 98


Some are duplicates. I wrote this program in Python.

for k in range(1,100):
for j in range(1,100):
for l in range(1,100):
if(k**3 + j**3 == l**2):
print k, j, l


It's crude but you might tinker with it to see what else you churn up.

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It may be of help to consider that $x^3 + y^3 = ( x + y ) \cdot ( x^2 - xy + y^2)$, so one could start by looking for values of $x$ and $y$ for which those factors are equal (and then for values where one factor "completes a square" with the other).

ADDENDUM: I had a little time to think more on this during my snowy walk home... We can show that if we set $x = y$ , the two factors are $2x \cdot x^2$, which only gets us the (2, 2, 4) family of solutions [apart from $(0,0,0)$].

In fact, if we set $y = kx$ ($k$ integral), the factors are $[k + 1]x \cdot [k^2 - k + 1]x^2$, so we have $k + 1 = k^2 - k + 1 \Rightarrow k = 0, 2$ if we require the two factors $( x + y )$ and $( x^2 - xy + y^2)$ to be equal. [Just noticed that $k = 2$ will produce, for instance, the $(1,2,3)$ family.] So almost all the solutions are ones for which one of these factors contains a divisor just once that also appears in the other factor just once. An example is $(4,8,24)$, for which the factors are $(4+8) \cdot (4^2 - 4 \cdot 8 + 8^2) = ( 2^2 \cdot 3 ) \cdot (3 \cdot 4^2)$ . This ought to help narrow the search somewhat.

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In view of Marty Cohen's answer, I realize I failed to consider rational possibilities $ny = mx$ with both $m$ and $n$ integral. More searching fun... –  RecklessReckoner Apr 23 '13 at 2:00
Can you use your result to show that (a^3+b^3)(c^3+d^3) not equal to f^3+g^3; that is, the product is not closed in form? a, b, c, d, f, and g are positive integers. The product is not closed in form, otherwise, the cubic equation would have a solution z=a^3+b^3. –  Obinly Apr 23 '13 at 23:50

I think I have found a rather general solution. Let us choose any co-prime positive integers $u$ and $v$. From $u^3+v^3=(u+v)(u^2-uv+v^2)$, we form $[u(u^3+v^3)]^3+[v(u^3+v^3)]^3=[(u^3+v^3)^2]^2$. In this way, we can find infinitely many integer solutions. However, $(1,2,3)$ is the only co-prime integer solution triplet.

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There's $(2,2,4)$ lurking here too.

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If $n$ and $m$ are relatively prime positive integers, there are infinitely many positive integral solutions to $x^n + y^n = z^m$.

Proof: there are positive integers $u$ and $v$ such that $un - vm = -1$, or $un+1 = vm$. Let $x = pa^u$, $y = q a^u$, and $z = a^v$. Then we want $(p^n+q^n)a^{un} = a^{vm} = a^{un+1}$, or $p^n+q^n = a$.

So, from $n$ and $m$ get $u$ and $v$. Then choose any $p$ and $q$, and compute $a = p^n+q^n$. Then get $x$, $y$, and $z$ as above ($x = pa^u$, $y = q a^u$, and $z = a^v$).

For this case, with $n=3$ and $m=2$, $u = 1$ and $v = 2$, so choose $p$ and $q$ and let $a = p^3+q^3$ $x = p a$, $y = qa$, and $z = a^2$.

As a check, $x^3+y^3 = (pa)^3 + (qa)^3 = (p^3+q^3)a^3 = a^4$ and $z^2 = (a^2)^2 = a^4$.

Of course, if you want $x$ and $y$ to be relatively prime, it is harder.

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...and so taking $\,n=m>2\,$ we have a counter example to FLT...oh, my god! ...:P) –  DonAntonio Apr 23 '13 at 2:43
@DonAntonio: The first line says "If $n$ and $m$ are relatively prime positive integers, ...". So how can $m=n$? –  John Bentin Apr 23 '13 at 5:56
@JohnBentin, thanks. For a moment I thought we had a counter- example....oh, just kidding. –  DonAntonio Apr 23 '13 at 10:36