Established bounds

---

Current bounds

The published bounds are

$$(2/\sqrt3+o(1))\sqrt N\le M(N)\le(2+o(1))\sqrt N.$$

All sum conditions include diagonal pairs. The upper bound is the trivial bound printed as display (37) by Erdős–Freud (1991), p. 204; they state without proof that the coefficient $2$ can be replaced by $1.98$.

Lower bound

Put $m=\lfloor N/3\rfloor$, choose an ordinary Sidon set $B\subset[1,m]$ of size $(1+o(1))\sqrt m$, and put $A=B\cup(N-B)$. The three kinds of sums lie in

$$[2,2m],\qquad[N-m+1,N+m-1],\qquad[2N-2m,2N-2],$$

respectively. These ranges are disjoint, including when $3\mid N$. Within either copy, uniqueness follows from the Sidon property. A mixed sum is $N+b-b'$; every nonzero difference in a Sidon set has a unique ordered representation. Only the sum $N$ repeats, with $|B|$ representations.

For the reflected construction, see Erdős–Freud (1991), source record, pp. 203–204. For the ordinary Sidon asymptotic, see Theorem 5 of source record, pp. 10–11. It can also be obtained from the following standard construction. For a prime $p$, a primitive root $g$, and $t\in\mathbb Z_{p-1}$, choose $a_t\in\mathbb Z_{p(p-1)}$ with $a_t\equiv t\pmod{p-1}$ and $a_t\equiv g^t\pmod p$. The sum of two such residues determines the sum and product of their second coordinates in $\mathbb F_p$, hence the unordered pair. This is a modular Sidon set of size $p-1$. Choosing a prime $p\le\sqrt m$ with $p\sim\sqrt m$ and taking integer representatives gives the claimed lower bound for all large $m$.