Why LTI system cannot generate new frequencies?
One of the definitive features of LTI systems is that they cannot generate any new frequencies which are not already present in their inputs.
One way to see why this is so, comes by observing the output's Fourier transform $Y(\omega) = H(\omega)X(\omega)$
$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \longleftrightarrow Y(\omega)=X(\omega)H(\omega),$$ holds only when the impulse response $h(t)$  exists and it will exist only when the system is LTI.
From a little thought, guided by a simple graphical plot, and using the convolution-multiplication property above, one can see that the frequency region of support $R_y$ (set of frequencies for which $Y(\omega)$ is non-zero), of the output $Y(\omega)$ is given by the intersection of the regions of support $R_x$ and $R_h$ of the inputs $X(\omega)$ and frequency response $H(\omega)$ of the LTI system:
$$R_y = R_x \cap R_h$$
And from set algebra we know that if $A = B \cap C$ then $A \subset B$ and $A \subset C$ . That is, an intersection is always less or equivalent to what are being intersected. Therefore, the region of support for $Y(\omega)$ will be less than or at most equal to the support of $X(\omega)$. Hence no new frequencies will be observed at the output.
Since this property is a necessary condition for being an LTI system, any system that fails to posses it, therefore, cannot be LTI.