Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $

Question

Consider the equation $$ x^3+y^3+z^3=(x+y+z)^3 $$ for triples of integers $(x, y, z) $.

I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $.

Are there more solutions?

Answer 1

$$(x+y+z)^3-(x^3+y^3+z^3)=3(x+y)(y+z)(z+x)$$

so the only solutions are the ones the OP observed and their cyclically symmetric counterparts. There's no essential number theory going on here, just an algebraic identity.

Answer 2

Too long for a comment. Interestingly, for $k>3$, there are non-trivial solutions to,

$k=4;\quad x_i = -2634, 955, 1770, 5400$: (Jacobi-Madden equation)

$$x_1^4+x_2^4+x_3^4+x_4^4 = (x_1+x_2+x_3+x_4)^4$$

$k=5;\quad x_i = - 3, - 54,24,28,67 $:

$$x_1^5+x_2^5+\dots+x_5^5 = (x_1+x_2+\dots+x_5)^5$$

$k=6;\quad x_i = -4170, -3187, -888, 1854, 3300, 3936, 4230$:

$$x_1^6+x_2^6+\dots+x_7^6 = (x_1+x_2+\dots+x_7)^6$$

$k=7;\quad x_i = -230, -353, -625, -673, 184, 443, 556$:

$$x_1^7+x_2^7+\dots+x_7^7 = (x_1+x_2+\dots+x_7)^7$$

and longer ones for $k = 8,9,10$ (See April 6 update).

P.S. The cases $k=4,5$ involve elliptic curves, hence there are an infinite number of co-prime solutions.

Answer 3

For such equations :

$$(x+y)^3+x^3+y^3+z^3=(x+y+z)^3$$

You can write the formula.

$$x=3p^2+18ps-s^2$$

$$y=15p^2-6ps-5s^2$$

$$z=3p^2-6ps+7s^2$$