the only values of for which these two methods do not find expansions in three or fewer terms are those congruent to 1 mod 24. Polynomial identities listed by provide three-term Egyptian fractions for whenever is one of: 2 mod 3 (above), 3 mod 4, 2 or 3 mod 5, 3, 5, or 6 mod 7, or 5 mod 8. Combinations of Mordell's identities can be used to expand for all except possibly those that are 1, 121, 169, 289, 361, or 529 mod 840. The smallest prime that these identities do not cover is 1009. By combining larger classes of modular identities, Webb and others showed that the natural density of potential counterexamples to the conjecture is zero: as a parameter goes to infinity, the fraction of values in the interval . that could be counterexamples tends to zero in the limit. Nonexistence of identities If it were possible to find solutions such as the ones above for enough different moduli, forming a complete covering system of congruences, the problem would be solved. However, as showed, a polynomial identity that provides a solution for values of congruent to mod can exist only when is not congruent to a square modulo . (More formally, this kind of identity can exist only when is not a quadratic residue modulo .) For instance, 2 is a non-square mod 3, so Mordell's result allows the existence of an identity for congruent to 2 mod 3. However, 1 is a square mod 3 (equal to the square of both 1 and 2 mod 3), so there can be no similar identity for all values of that are congruent to 1 mod 3. More generally, as 1 is a square mod for all , there can be no complete covering system of modular identities for all , because 1 will always be uncovered. Despite Mordell's result limiting the form of modular identities for this problem, there is still some hope of using modular identities to prove the Erdős–Straus conjecture. No prime number can be a square, so by the Hasse–Minkowski theorem, whenever is prime, there exists a larger prime such that is not a quadratic residue modulo . One possible approach to proving the conjecture would be to find for each prime a larger prime and a congruence solving the problem for congruent to mod . If this could be done, no prime could be