Diophantine cubic equations

Question

Specifically I mean the equation: $$x^3+y^3=u^3+v^3+w^3.$$ It is relatively easy to find a solution (in this case) by guessing: $$7^3+8^3=1^3+5^3+9^3.$$ However, I don't know whether that solution is unique. So my question is if there is some algorithm to deal with that kind of equations and, more importantly, if there is not solution, there is one solution, or there are several solutions. Thanks for helping!

Answer 1

Some solutions: $$ \eqalign{13^3 + 14^3 &= 1^3 + 3^3 + 17^3\cr 7^3 + 8^3 &= 1^3 + 5^3 + 9^3\cr 7^3 + 21^3 &= 1^3 + 14^3 + 19^3\cr 17^3 + 18^3 &= 1^3 + 14^3 + 20^3\cr 7^3 + 9^3 &= 2^3 + 4^3 + 10^3\cr 14^3 + 16^3 &= 2^3 + 10^3 + 18^3\cr 5^3 + 15^3 &= 3^3 + 9^3 + 14^3\cr 2^3 + 19^3 &= 3^3 + 14^3 + 16^3\cr 10^3 + 21^3 &= 3^3 + 15^3 + 19^3\cr 1^3 + 7^3 &= 4^3 + 4^3 + 6^3\cr 14^3 + 18^3 &= 4^3 + 8^3 + 20^3\cr 7^3 + 16^3 &= 4^3 + 10^3 + 15^3\cr 10^3 + 21^3 &= 4^3 + 13^3 + 20^3\cr 11^3 + 17^3 &= 5^3 + 14^3 + 15^3\cr 12^3 + 20^3 &= 5^3 + 14^3 + 19^3\cr 7^3 + 11^3 &= 6^3 + 9^3 + 9^3\cr 10^3 + 14^3 &= 6^3 + 11^3 + 13^3\cr 3^3 + 15^3 &= 7^3 + 11^3 + 12^3\cr 5^3 + 19^3 &= 7^3 + 12^3 + 17^3\cr 2^3 + 14^3 &= 8^3 + 8^3 + 12^3\cr 3^3 + 21^3 &= 10^3 + 15^3 + 17^3\cr 7^3 + 17^3 &= 11^3 + 12^3 + 13^3\cr 3^3 + 21^3 &= 12^3 + 12^3 + 18^3\cr 14^3 + 22^3 &= 12^3 + 18^3 + 18^3\cr 5^3 + 22^3 &= 13^3 + 14^3 + 18^3\cr 3^3 + 26^3 &= 14^3 + 19^3 + 20^3\cr }$$

If $(x,y,u,v,w)$ is a solution, then so is $(tx,ty,tu,tv,tw)$ for any positive integer $t$, so there are infinitely many solutions. But I suspect there are infinitely many even if you require $\text{gcd}(x,y,u,v,w)=1$.

EDIT: For example, given the solution $7^3 + 8^3 = 1^3 + 5^3 + 9^3$, we might look for solutions $(x,y,u,v,w)$ such that $x=7 u$ and $v = 5 u$. Then $u,w,y$ must satisfy $217 u^3 + y^3 - w^3 = 0$. Letting $a = u/w$ and $b = y/w$, we are looking for rational solutions to $217 a^3 + b^3 - 1 = 0$. That turns out to be an elliptic curve. After turning it into Weierstrass form $A^3 - B^3 - 1271403/4 = 0$, Sage tells me this has rank $2$. Therefore there should be infinitely many solutions of the given form. For example, the generators, according to Sage, are $A = 273/4$, $B = 63/8)$, which corresponds to $a=26/125, b=-123/125$, and $A = 93, B=1395/2$, which corresponds to $a=1/6$, $b = -1/6$.

Answer 2

Above equation shown below:

$x^3+y^3=u^3+v^3+w^3$ --------(1)

Equation (1) has parametric solution & is shown below;

$x=8(p-1)^2$

$y=(7p^2-16p+8)$

$u=(3p-2)(3p-4)$

$v=(p-2)(5p-4)$

$w=(p)^2$

For, $p=3$ we get:

$32^3+23^3=35^3+11^3+9^3$

Answer 3

$$X_{1}^{3}+X_{2}^{3}+X_{3}^{3}=X_{4}^{3}+X_{5}^{3}$$

I kind of wrote 5 parametric solution.

$$X_{1}=6ca^{2}(k-p)(t-p)$$

$$X_{2}=c^{3}p^{2}+(3a^{3}t-(3a^{3}+2c^{3})k)p+c^{3}k^{2}+3a^{3}kt-3a^{3}t^2$$

$$X_{3}=c^{3}p^{2}+((3a^{3}-2c^{3})k-3a^{3}t)p+c^{3}k^{2}-3a^{3}kt+3a^{3}t^{2}$$

$$X_{4}=(c^{3}-6a^{3})p^{2}+(9a^{3}t+(3a^{3}-2c^{3})k)p+c^{3}k^{2}-3a^{3}kt-3a^{3}t^{2}$$

$$X_{5}=(c^{3}+6a^{3})p^{2}-(9a^{3}t+(3a^{3}+2c^{3})k)p+c^{3}k^{2}+3a^{3}kt+3a^{3}t^{2}$$

For the case when $k=p$ or $t=p$.... get the formula for 4 cubes...

Answer 4

$x^3+y^3=u^3+v^3+w^3\tag{1}$
We can get infinitely many integer solutions from a known solution.
Let $[1,-1,1,-1,0]$ is a known solution for equation $(1)$.
Substitute $x=mt+1, y=t-1, u=t+1, v=nt-1, w=t+0$ to equation $(1)$, then we get $$(m^3-1-n^3)t^3+(3m^2-6+3n^2)t^2+(3m-3n)t=0$$ Let $m = n$ then $t = 6n^2-6$.
Then we get a parametric solution below.
$$x = 6n^3-6n+1$$ $$y = 6n^2-7$$ $$u = 6n^2-5$$ $$v = 6n^3-6n-1$$ $$w = 6(n-1)(n+1)$$ n is arbitrary.
Thus, we get infinitely many integer solutions from a known solution.

Example: $n=2..5,$ $[x,y,u,v,w]=[37, 17, 19, 35, 18], [145, 47, 49, 143, 48], [361, 89, 91, 359, 90], [721, 143, 145, 719, 144]$