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$(x,y,z)=(n+3,3,2)$ solves your example. Your form is of the form $x\cdot a(y,z) + b(y,z)=n$. If you can choose $y,z$ such that $a(y,z)=1$ you have a solution for all $n$ large enough.
$(x,y,z)=(n+3,3,2)$ solves your example.
$(x,y,z)=(n+3,3,2)$ solves your example. Your form is of the form $x\cdot a(y,z) + b(y,z)=n$. If you can choose $y,z$ such that $a(y,z)=1$ you have a solution for all $n$ large enough.
