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If $2(x^3+y^3+z^3) = 3(x+y+z)^2$then
find the maximum value of $x+y+z$.
given that $x,y,z$ are all non negative integers.

The Z3 solver does not find an objective value larger than $2$:

from z3 import *



X, Y, Z = Ints('X Y Z')



s = Optimize()



s.add(X >= 0, X <= Y, X < 10)

s.add(Y >= 0, Y <= Z, Y < 10)

s.add(Z >= 0, Z < 10)

s.add(2*(X**3+Y**3+Z**3) == (X+Y+Z)**2)



obj = Sum(X+Y+Z)

s.maximize(obj)



print(s.check())

try:

  print(s.model())

except Z3Exception as ex:

  print(ex)

I limited the value ranges and imposed an $ X le Y le Z$ ordering to get a fast answer.

Result:

sat

[Y = 1, Z = 1, X = 0]

WLOG we may assume $xgeq ygeq zgeq0$. It is easy to see that $(x,y,z)=(1,1,0)$ is a solution. Let's treat $x,y$ as constant and consider both the LHS and RHS as functions of $z$ only. Clearly, the LHS is a cubic polynomial which is monotonously increasing on $[0,infty)$, and the RHS is a quadratic polynomial which also increases monotonously on $[0,infty)$. If we plug in $z=1$, we have $mathrm{LHS}=6$ and $mathrm{RHS}=9$. If we plug in $z=2$, then $mathrm{LHS}=20$ and $mathrm{RHS}=16$, and the LHS has already "outgrowed" the RHS. Indeed this will always happen regardless of the values of $x,y$ you choose, and a similar argument (by treating $y,z$ as constant and $x$ as the variable, for example) shows that increasing $x,y$ from $(1,1,0)$ will only make the LHS "outgrow" the RHS quicker. Hence, $(1,1,0)$ and $(1,1,1)$ are the only two possible solutions to satisfy $mathrm{LHS}leqmathrm{RHS}$, and since $(1,1,1)$ doesn't work, the unique solution to the equation is $(1,1,0)$.