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

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4  
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
    
A trivial (perhaps interesting)observation: Suppose we have two solutions to: A^3 + B^3 = x^2 C^3 + D^3 = y^2 Then there exists no number A such that A^3 = y^2 – x^2. Is this correct? V. V. Raman –  user144775 Apr 21 at 21:55

11 Answers 11

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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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

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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You can work in $\mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{3}}.$ There are two essential cases to deal with. In both cases, you can easily reduce to considering the case that $x,y,z$ are pairwise coprime ( you can afterwards multiply $x$ and $y$ by $d^{2}$ and $z$ by $d^{3}$ for some integer $d$). The case when $3$ does not divide $z$ is easier.

In that case, you can assume that $x + y = a^{2}$ for some integer $a$ and that $x + \omega y = (b + \omega c)^{2}$ for integers $b$ and $c.$ Then you can see that $x^{3} + y^{3} = (a(b^{2}-bc +c^{2}))^{2}.$ Note that the second equation gives $x = b^{2}-c^{2}$ and $y = 2bc - c^{2}$ since $1 + \omega + \omega^{2} = 0.$ Hence we additionally need the integers $a,b,c$ to satisfy $a^{2} + 3c^{2} = (b+c)^{2},$ but this is easy to arrange.

In the (still pairwise coprime case that $3$ divides $z,$ you must have $x + y = 3^{2m-1}a^{2}$ for some integer $a$ coprime to $3$, and $x + \omega y = (1- \omega)(b + \omega c)^{2}$ for some integers $b$ and $c$ with $b+c$ not divisible by $3$.

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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

equations $X^3+Y^3=Z^2$ Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.

$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$

$Y=2b(a+b)(3a^2+b^2)c^2$

$Z=3(a+b)^2(3a^2+b^2)^2c^3$

And more.

$X=2b(b-a)(3a^2+b^2)c^2$

$Y=2b(b+a)(3a^2+b^2)c^2$

$Z=4b^2(3a^2+b^2)^2c^3$

If we decide to factor $X^3+Y^3=qZ^2$

For a compact notation we replace :

$a=s(2p-s)$

$b=p^2-s^2$

$t=p^2-ps+s^2$

then:

$X=qb(a+b)c^2$

$Y=qa(a+b)c^2$

$Z=qt(a+b)^2c^3$

And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$

by the way $a$ May appear as a factor in the decision and. Then the solutions are of the form::

$X=qa(2p-3as)sc^2$

$Y=q(p-2as)pc^2$

$Z=q(p^2-3aps+3a^2s^2)c^3$

If we change the sign : $Y^3-X^3=qZ^2$

Then the solutions are of the form:

$X=qa(2p+3as)sc^2$

$Y=q(p+2as)pc^2$

$Z=q(p^2+3aps+3a^2s^2)c^3$

Another solution of the equation: $X^3+Y^3=qZ^2$

$p,s$ - integers asked us.

To facilitate the calculations we make the change. $a,b,c$

If the ratio is as follows : $q=3t^2+1$

$b=2q(q+2\mp{6t})p^2+6p(t\mp1)ps+(q-1\mp{3t})s^2$

$c=6q(q-2(1\pm{t}))p^2+6q(t\mp1)ps+3(1\mp{t})s^2$

$a=12q(1\mp{t})p^2+6(4t\mp{q})ps+3(1\mp{t})s^2$

If the ratio is as follows: $q=t^2+3$

$b=3(q-1)(1\pm{t})s^2+2(3\pm{(q-1)t})ps+(1\pm{t})p^2$

$c=3(6-(q-1)(q-3\mp{t}))s^2+6(1\pm{t})ps+(q-3\pm{t})p^2$

$a=3(6-(q-1)(1\mp{t}))s^2+6(1\pm{t})ps+(1\pm{t})p^2$

Then the solutions are of the form:

$X=2c(c-b)$

$Y=(c-3b)(c-b)$

$Z=3a(c-b)^2$

Then the solutions are of the form:

$X=2(c-b)c$

$Y=2(c+b)c$

$Z=4ac^2$

If the ratio is as follows : $q=t^2+3$

$c=6(q-4)(2\pm{t})p^2+4(6\pm{(q-4)t})ps+2(2\pm{t})s^2$

$b=3(24-(q-4)(q-3\mp{2t}))p^2+12(2\pm{t})ps+(q-3\pm{2t})s^2$

$a=3(24-(q-4)(4\mp{2t}))p^2+12(2\pm{t})ps+2(2\pm{t})s^2$

If the ratio is as follows $q=3t^2+4$

$c=q(-q+7(4\mp{3t}))p^2+6q(t\mp{1})ps+(q-4\mp{3t})s^2$

$b=3q(2q-7(1\pm{t}))p^2+6q(t\mp{1})ps+3(1\mp{t})s^2$

$a=21q(1\mp{t})p^2+6(7t\mp{q})ps+3(1\mp{t})s^2$

Then the solutions are of the form :

$X=2(3c-2b)c$

$Y=2(3c+2b)c$

$Z=12ac^2$

Then the solutions are of the form :

$X=(2b-c)b$

$Y=(2b+c)b$

$Z=2ab^2$

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Well you can and draw another formula. $x^3+y^3=z^2$

$x=(b^2-a^2)(b^2+2ba-2a^2)c^2$

$y=(b^2-a^2)(2b^2-2ab-a^2)c^2$

$z=3(b^2-a^2)^2(a^2-ab+b^2)c^3$

The most interesting thing there is that the formula that led, like should not give mutually simple solutions, but after sokrasheniya on common divisor can be obtained and are relatively prime solutions. This means that the formula itself describes as relatively prime so no. Coprime solutions - there are private solutions.

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