A nonlinear resistor has a bent/distorted \$I\$-\$V\$-curve. We can approximate its \$I\$-\$V\$-curve with a polynomial model such as:
$$I(V)=I_0 + \frac{1}{R_1}V + \frac{1}{R_2}V^2 + \frac{1}{R_3}V^3 + \cdots$$
If all the terms other than \$\frac{V}{R_1}\$ are zero, we end up with a linear resistor of value \$R_1\$.
Assume we measure this resistor in such an arrangement:
simulate this circuit – Schematic created using CircuitLab
The current through the nonlinear resistor is given by \$\frac{V_C}{R_C}\$ and the voltage across the nonlinear resistor is \$V_{NL}\$. As the resistor is nonlinear, both voltmeters will record harmonic distortion, even if the source is a pure sine and if \$R_C\$ is a perfectly linear resistor:
$$I_\text{res}(V)=I_0 + I_1 \sin(\omega_1 t+\phi_1) + I_2 \sin(\omega_2 t+\phi_2) + \cdots$$ $$V_\text{res}(V)=V_0 + V_1 \sin(\omega_1 t+\phi_1) + V_2 \sin(\omega_2 t+\phi_2) + \cdots$$
What is the relation between the complex Fourier components in the last two equations (e.g. \$I_1\angle\phi_1\$, \$I_2\angle\phi_2\$, ...) and the polynomial approximation of the \$I\$-\$V\$-curve using the coefficients \$R_1\$, \$R_2\$, ... ?
Clearly, there must be a relation as they describe the same phenomenon, namely the nonlinearity of the tested resistor.