It is possible to divide $\mathbb{N}$ is four disjoint sets $A_1,A_2,A_3,A_4$ such that if $x+2y=5z$, then $x,y,z$ are not in the same set: $$A_1=\{5^ab| b\equiv \pm 1 (\mathrm{mod}\ 5),\ a \mathrm{\ even}\}$$$$A_2=\{5^ab| b\equiv \pm 1 (\mathrm{mod}\ 5),\ a \mathrm{\ odd}\}$$$$ A_3=\{5^ab| b\equiv \pm 2 (\mathrm{mod}\ 5),\ a \mathrm{\ even}\}$$$$A_4=\{5^ab| b\equiv \pm 2 (\mathrm{mod}\ 5),\ a \mathrm{\ odd}\}$$ $x+2y=5z$ doesn't hold in any $A_i$ : the parity assures us that after dividing with $5^{\mathrm{min}( v_5(x),v_5(y))}$, the RHS will still be divisible by $5$; however, it is easy to see that we cannot have a multiple of $5$ in the LHS, thus the conclusion.