I am studying the equation
$$ x^m+y^m+z^m=(x+y)^m, $$
(and the related inequalities) where $x,y,z,m\in\mathbb{N}$ and $x,y,z,m>0$.
Is there anybody familiar with such equation, and/or knows where I can find some material about it? Thanks for your help!
This post is related to the conjecture exposed here A conjecture involving the equation $x^n+y^n+z^n= (x+y)^n$.
For $m=3$ there is a solution in small integers:
$$1^3 + 8^3 + 6^3 = 9^3 = (1+8)^3\quad\quad(1)$$
To find further solutions, we can transform the equation to an elliptic curve as below. Put $x=1$ (this is without loss of generality as any solution in integers is equivalent to a solution in rationals with $x=1$, and any such solution in rationals can be converted to one in integers by multiplying through by the cube of the least common multiple of the denominators):
$$1 + y^3 + z^3 = (1+y)^3$$
$$z^3 = 3y(1+y)$$
Put $t = y + \frac{1}{2}$:
$$z^3 = 3(t-\tfrac{1}{2})(t+\tfrac{1}{2}) = 3t^2 - \tfrac{3}{4}$$
$$t^2 = \frac{z^3}{3} + \frac{1}{4}\quad\quad(2)$$
From $(1)$, we have the solution to (2): $(t,z)=(\tfrac{17}{2},6)$. To use the elliptic curve $(2)$ to find new solutions we need one more initial solution. Surprisingly, perhaps, we can obtain a suitable second solution by rearranging $(1)$ as below (although it contains negatives it serves perfectly well as a starting point for finding further solutions in positives):
$$9^3 + (-1)^3 + (-6)^3 = 8^3$$
$$1 + (-\tfrac{1}{9})^3 + (-\tfrac{6}{9})^3 = (\tfrac{8}{9})^3\quad\quad(3)$$
Since $-\tfrac{1}{9}+\tfrac{1}{2}= \tfrac{7}{18}$ and $-\tfrac{6}{9}=-\tfrac{2}{3}$, this implies the solution to $(2)$: $(t,z)=(\tfrac{7}{18},-\tfrac{2}{3})$.
Letting $z=at+b$ be the straight line through the above two solutions to $(2)$ we have:
$$6 = \tfrac{17}{2}a + b$$ $$-\tfrac{2}{3} = \tfrac{7}{18}a + b$$
Subtracting:
$$6+\tfrac{2}{3} = (\tfrac{17}{2}-\tfrac{7}{18})a$$
$$\tfrac{20}{3} = \tfrac{146}{18}a$$
$$a = \tfrac{360}{438} = \tfrac{60}{73}$$
Hence:
$$b = 6 - (\tfrac{17}{2})(\tfrac{60}{73}) = 6 - \tfrac{510}{73} = -\tfrac{72}{73}$$
So we expect a third solution where $(2)$ meets the line $z = \frac{60}{73}t - \tfrac{72}{73}$. The calculation is tedious to write out in full but it can be shown that the solution is $(z,t) = (-\tfrac{7140}{8000},\tfrac{913}{8000})$. To check this, note that:
$$\frac{z^3}{3}+\tfrac{1}{4} = -\tfrac{7140^3}{3(8000^3)}+\tfrac{1}{4} = -\tfrac{357^3}{3(400^3)}+\tfrac{1}{4}= \tfrac{-15166431+16000000}{64000000}=-\tfrac{833569}{64000000}=\tfrac{913^2}{8000^2}=t^2$$
and:
$$\tfrac{60t}{73}-\tfrac{72}{73} = \tfrac{(60(913/8000))-72}{73} =\tfrac{54780-576000}{584000}=\tfrac{-521220}{584000}=-\tfrac{7140}{8000}=z$$
To obtain a solution in rationals of the original equation we have
$$y=t-\tfrac{1}{2}= \tfrac{913}{8000}-\tfrac{1}{2} = -\tfrac{3087}{8000}$$
Hence:
$$1^3 + (-\tfrac{3087}{8000})^3 + (-\tfrac{7140}{8000})^3 = (\tfrac{4913}{8000})^3$$
Multiplying through by $8000^3$ and rearranging:
$$4913^3 + 3087^3 + 7140^3 = 8000^3 = (4913+3087)^3$$
Further solutions may be obtained by finding further rational points on the elliptic curve $(2)$. For a general explanation of how to find rational points on elliptic curves see Silverman J H, An Introduction to the Theory of Elliptic Curves, especially section on The Geometry of Elliptic Curves pp 8ff.
I'm posting this as a separate answer, again for case $m=3$, as its reasoning is quite different to that of my earlier answer.
Suppose we have a solution in integers to the equation:
$$a^3 + b^3 = 9c^3\quad\quad(1)$$
Then we can infer a solution to:
$$x^3 + y^3 + z^3 = (x+y)^3\quad\quad(2)$$
since:
$$(a^3)^3 + (b^3)^3 + (3abc)^3 = (a^3)^3 + (b^3)^3 + 3a^3b^3(a^3+b^3) = (a^3 + b^3)^3\quad\quad(3)$$
We can also infer a further solution to (1) and therefore to (2) since, given (1):
$$(a(a^3 + 2b^3))^3 + (-b(2a^3 + b^3))^3 = 9(c(a^3 - b^3))^3\quad\quad(4)$$
Proof:
LHS of $(4) = a^{12} + 6a^9b^3 + 12a^6b^6 + 8a^3b^9 - 8a^9b^3 - 12a^6b^6 - 6a^3b^9 - b^{12}$
$\quad\quad\quad\quad= (a^{12} - 2a^9b^3 + a^6b^6) - (a^6b^6 - 2a^3b^9 + b^{12})$
$\quad\quad\quad\quad= (a^6 - b^6)(a^6 - 2a^3b^3 + b^6)$
$\quad\quad\quad\quad= (a^3 + b^3)(a^3 - b^3)(a^3 - b^3)^2$
$\quad\quad\quad\quad= 9(c(a^3 - b^3))^3 = \text{RHS of } (4)$.
The above is a special case of a proof of a more general proposition (replacing $9$ with any cube-free integer $N$) that may be found in (A).
Since we have a solution to $(1)$, namely:
$$1^3 + 2^3 = 9(1^3)$$
we can use the above to obtain a series of solutions to (2) which, with suitable changes of sign and rearrangement as necessary, yield solutions in positive integers. Successive solutions increase rapidly in size, the first three being:
First:
$$1^3 + 2^3 = 9(1^3)$$
$$\Rightarrow(1^3)^3 + (2^3)^3 + (3(1)(2)(1))^3 = ((1^3)+(2^3))^3$$
$$\Rightarrow 1^3 + 8^3 + 6^3 = (1+8)^3 = 9^3$$
Second:
$$17^3 + (-20)^3 = 9(7^3)$$
$$\Rightarrow(9(7^3))^3 + (17^3)^3 + (3(7)(17)(20))^3 = (20^3)^3$$
$$\Rightarrow3087^3 + 4913^3 + 7140^3 = 8000^3$$
Third:
$$(-36520)^3 + 188479^3 = 9(90391^3)$$
$$\Rightarrow(9(90391^3))^3 + (36520^3)^3 + (3(36520)(188479)(90391))^3 = (188479^3)^3$$
$$\Rightarrow6646883738818240^3 + 48707103808000^3 + 1866552387462840^3 = (6646883738818240 + 48707103808000)^3 = 6695590842626240^3$$
(A) Reference: Gordon R A (2018), Rational Arc Length The Mathematical Gazette Vol 102 No. 554 pp 210-225 (see especially pp 213-4).