 Solution

Let $a=\lambda /6$. Within terms already suppressed by ${\Lambda }^{-2}$, the leading equation of motion gives $\square \varphi =-m^2\varphi -a{\varphi }^3$. Equality here means equivalence under a perturbative field redefinition, together with integration by parts, rather than a pointwise identity inside arbitrary off shell correlation functions. HenceFor the second relation, factor the difference as $(\square \varphi -m^2\varphi -a{\varphi }^3)(\square \varphi +m^2\varphi +a{\varphi }^3)$, which is explicitly a redundant operator. Ifthe reduced correction isThe first two coefficients shift the already present mass and quartic coupling constant. After these shifts,This is the equation-of-motion reduction of scalar operators through $O({\Lambda }^{-2})$; terms of $O({\Lambda }^{-4})$ are outside that truncation. The lower-dimensional shifts should not be silently dropped when $m\ne 0$.