When does $x^3+y^3=kz^2$?

Question

For which integers $k$ does $$ x^3+y^3=kz^2 $$ have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?

Answer 1

Answer (Version III)

It turns out if we allow $x$, $y$ to be negative, all $k$ is possible!
For any integer $t$, define

$$ \begin{cases} x_o(t) &= 1-3t^2\\ y_o(t) &= 3t^2 + 6t + 2\\ z_o(t) &= 3(3t^2+3t+1)\\ \end{cases} \quad\text{ and }\quad \begin{cases} x_e(t) &= 12t^2 + 12t - 1\\ y_e(t) &= -12t^2 + 12t + 1\\ z_e(t) &= 6(12t^2+1) \end{cases} $$ We have $$ x_o(t)^3 + y_o(t)^3 = (2t+1) z_o(t)^2 \quad\text{ and }\quad x_e(t)^3 + y_e(t)^3 = 2t z_e(t) $$

Furthermore, $$ \begin{cases} (2t+3)x_o(t) + (2t-1)y_o(t) &= 1\\ (t-1)x_e(t) + (t+1)y_e(t) &= 2 \end{cases} \quad\implies\quad \begin{cases} \gcd(x_o(t),y_o(t)) = 1\\ \gcd(x_e(t),y_e(t)) = 1 \text{ or } 2\\ \end{cases} $$

Since $x_e(t), y_e(t)$ are odd numbers, the possibilities that $\gcd(x_e(t),y_e(t)) = 2$ has been ruled out.
As a result, all the $( x_o, y_o )$ and $(x_e, y_e)$ are co-prime solutions for corresponding Pell-Fermat equation.

Answer (Version I)

(obsoleted by findings in Version II, keep here for historical reference)

This is just another set of random data points. The main result is

If the Diophantine equation $b^2 - 3k a^2 = -2$ has a non-trivial solution $(a,b)$ for positive integer $k$, then $$\begin{cases} x &= 6c^2 + 6c + 1\\ y &= 3c^2 + 6c + 2\\ z &= 3ab (3c^2 + 3c + 1) \end{cases} \quad\text{ with }\quad c = k a^2-1 $$ will be a solution for the Pell-Fermat equation $k z^2 = x^3 + y^3$.

Since $$(2c+3)(6c^2 + 6c + 1) - (4c+2)(3c^2 + 6c + 2) = -1$$

The pair of $x,y$ generated in this manner is always co-prime.

By brute force searching, the Diophantine equation $b^2 - 3ka^2 = -2$ does have non-trivial solutions for following list of $k \le 100$.

$$1,2,6,9,17,18,22,34,38,41,54,57,66,81,82,86,89,97$$

As a result, the corresponding Pell-Fermat equation does have a solution for these $k$.
Following is a table showing one solution for each $k$.

$$\begin{array}{rcrcrcr} z^2 &\times& k &=& x^3 &+& y^3\\ \hline 3^2 &\times& 1 &=& 1^3 &+& 2^3\\ 42^2 &\times& 2 &=& 13^3 &+& 11^3\\ 1092^2 &\times& 6 &=& 181^3 &+& 107^3\\ 3255^2 &\times& 9 &=& 433^3 &+& 242^3\\ 17157^2 &\times& 17 &=& 1633^3 &+& 866^3\\ 15492906^2 &\times& 18 &=& 156493^3 &+& 78731^3\\ 33288^2 &\times& 22 &=& 2773^3 &+& 1451^3\\ 101010^2 &\times& 34 &=& 6733^3 &+& 3467^3\\ 100761696^2 &\times& 38 &=& 699733^3 &+& 350891^3\\ 162393^2 &\times& 41 &=& 9841^3 &+& 5042^3\\ 591636585540^2 &\times& 54 &=& 256119733^3 &+& 128079467^3\\ 373503^2 &\times& 57 &=& 19153^3 &+& 9746^3\\ 540582^2 &\times& 66 &=& 25741^3 &+& 13067^3\\ 22216951933755^2 &\times& 81 &=& 3287747233^3 &+& 1643943842^3\\ 44651886469302^2 &\times& 82 &=& 5257492813^3 &+& 2628835211^3\\ 1052688^2 &\times& 86 &=& 43861^3 &+& 22187^3\\ 847778841^2 &\times& 89 &=& 3844801^3 &+& 1924802^3\\ 1424787^2 &\times& 97 &=& 55873^3 &+& 28226^3\\ \end{array}$$

Answer 2

If $k=z$, there are no solution by Fermat's Last Theorem.

When $k=1$ a solution can be $x=1$, $y=2$, $z = 3$, because $1^{3} + 2^{3} = 3^{2}$

Another $k=38$ have a solution can be $x =3$, $y=5$, $z=2$ and there are lots of $k$ can be solution of that equation, but now I don't know any technique for count how many or what $k$ give a solution.

EDIT:

I found the following solution, I think it isn't the best but:

\begin{equation} x^{3} + y^{3} = (x+y)[(x-y)^{2}+xy] \end{equation}

we can think that $(x+y) = z^{2}$ and $k =[(x-y)^{2}+xy]$.

now using $xy = \frac{1}{4}[(x+y)^2 - (x-y)^{2}]$

we can found $3$ solutions,

$1) k =[(x-y)^{2}+xy] \iff x+y$ is a square;

$2) k =[\frac{3}{2}(x-y)^{2}+\frac{(x+y)^2}{2}] \iff \frac{x+y}{2}$ is a square;

$3) k =[3(x-y)^{2}+(x+y)^2] \iff \frac{x+y}{4}$ is a square;

Answer 3

Well you can and draw another formula. $$x^3+y^3=kz^2$$ $$x=k(b^2-a^2)(b^2+2ba-2a^2)c^2$$ $$y=k(b^2-a^2)(2b^2-2ab-a^2)c^2$$ $$z=3k(b^2-a^2)^2(a^2-ab+b^2)c^3$$

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.

the equation: $$(X+1)^3-X^3=Y^2$$

Solutions can be written if we use the solutions of Pell's equation: $p^2-3s^2=1$

Then the solution can be written: $$X=\pm2ps-p^2$$ $$Y=p^2\mp3ps+3s^2$$

If we make the change: $$q=2ps\pm(p^2+3s^2)$$ $$t=4ps\pm(p^2+3s^2)$$

Then the solutions are of the form: $$Y=t^2-tq+q^2$$ $$X=t(t-2q)$$

If we make the change: $$t=3ps\pm(p^2+3s^2)$$ $$q=4ps\pm(p^2+3s^2)$$

Then the solutions are of the form: $$X=-4t(t+q)$$ $$Y=4t^2+6tq+3q^2$$

There are formulas, but I have to rewrite no time. You can see it there.

http://www.artofproblemsolving.com/blog/98941

Answer 4

$\qquad\qquad$ Too long for a comment: Positive coprime integers x and y less than $6,000$ :

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=1$

$\qquad\qquad\qquad\qquad$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=2$

$\qquad\qquad\qquad\qquad$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=3$

$\qquad\qquad\qquad\qquad$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=4$

$\qquad\qquad\qquad\qquad$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=5$

$\qquad\qquad\qquad\qquad$

Answer 5

A question to achille hui

"Actually, this result is a byproduct of a more general approach which study the equations over the Eisenstein integers Z[e 2πi/3 ] . For each k , there are finitely many ways to factorize it over Z[e 2πi/3 ] and for each factorization, the problem can be reduced to the problem of finding a solution for a Diophantine equation on 3 variables."

Achille, do you have any references about this approach? Any published papers? Do you mean QUADRATIC Diophantine equation of 3 variables? Thank you, Vlad

Answer 6

Consider the equation: $$(a+b)^3+(a-b)^3=2a(a^2+3b^2)$$ Now let $a^2+3b^2=c^2$. We want to find a general equation for this. Note that $$(m-n)^2+2mn=(m+n)^2$$ Therefore, substitute $m\mapsto m^2$ and $n\mapsto n^2k$ then $$(m^2-k\cdot n^2)^2+k(2mn)^2=(m^2+k\cdot n^2)^2$$ Let $k=3$, and it follows that $a=m^2-3n^2$ and $b=2mn$, given that $a^2+b^2=(m^2+3n^2)^2$. $$\therefore (m^2-3n^2+2mn)^3+(m^2-3n^2-2mn)^3=2(m^2-3n^2)(m^2+3n^2)^2$$ We could also substitute $m\to 3m^2$ and $n\mapsto n^2$ to get $$(3m^2-n^2+2mn)^3+(3m^2-n^2-2mn)^3=2(3m^2-n^2)(3m^2+n^2)^2$$