Let a polynomial function:
f(x)=$x^n+a_{n−1}x^{n−1}+⋯+a_2x^2+a_1x+a_0$
Where n is positive integer, $n \geq 3$ and $a_0, a_1, a_2, ..., a_n$ are constant coefficients $\in Z$ and $f(x)=0$ has only solution $p \in Z$. Then has only $x=z,y =p or y=z,x =p or x=y=z=p$ with $x,y,z \in Z$ can satisfy the equation:
f(x)=$x^n+a_{n−1}x^{n−1}+⋯+a_2x^2+a_1x+a_0$
Where n is positive integer, $n \geq 3$ and $a_0, a_1, a_2, ..., a_n$ are constant coefficients $\in Z$ and $f(x)=0$ has only solution $p \in Z$. Then has only $x=z,y =p or y=z,x =p or x=y=z=p$ with $x,y,z \in Z$ can satisfy the equation:
$f(x)+f(y)=f(z)$
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