Let $L$ denote a lattice. If no sublattice of $L$ is shaped like the pentagon $N_5$, we call $L$ modular. Supposing furthermore that no sublattice of $L$ is shaped like the diamond $M_3$, we call $L$ distributive.
What do we call a lattice that does not have a sublattice the shape of the diamond $M_3$, but which might have a sublattice the shape of $N_5$? Furthermore, is there a nice "algebraic" characterization of this property, like there is for modularity and distributivity?
Let $\mathbf K$ be the class of all lattices not containing $M_3$ (the $5$-element nondistributive modular lattice) as a sublattice; in other words, lattices in which every modular sublattice is distributive. It is easy to see that $\mathbf K$ can be characterized as the class of all lattices satisfying the following sentence $\varphi$: $$\forall u\forall v\forall a\forall b\forall c[(ab=ac=bc=u)\wedge(a+b=a+c=b+c=v)\rightarrow(u=v)]$$ or equivalently $$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(ab=a+b)]$$ or $$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(a=b)].$$
Inasmuch as $\varphi$ is a universal Horn sentence, it follows that $\mathbf K$ is closed under taking sublattices and direct products.
On the other hand, $\mathbf K$ can not be characterized by identities, and is not closed under taking homomorphic images. Bjarni Jónsson [Sublattices of a free lattice, Canad. J. Math. 13 (1961), 256-264, Lemma 2.6(i)] observed that, as a corollary of P. M. Whitman's work, elements $u,a,b,c$ of a free lattice satisfy the condition: if $u=ab=ac$, then $u=a(b+c)$. It follows that $M_3$ is not a sublattice of a free lattice, i.e, the class $\mathbf K$ contains all free lattices. Thus every lattice is a homomorphic image of a member of $\mathbf K$, and every identity which holds in all $M_3$-free lattices is a consequence of the lattice axioms. In view of this, there is not likely to be any "nicer" algebraic characterization of the $M_3$-free lattices than the one in the previous paragraph.