Bloom 2022 egyptian fractions

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theorem_1: The Erdős–Straus conjecture holds if and only if every prime lies in one of the congruence classes -a/c mod 4acd-1 or -(4c^2d+1)/k mod 4cd.

theorem_3: The survey's restatement of Elsholtz's bound: for m > k >= 3 the number of n <= N for which m/n is not a sum of k unit fractions is at most N exp(-c (log N)^(1 - 1/(2^(k-1) - 1))), which for m = 4, k = 3 is Vaughan's bound.

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Bloom, Thomas F. and Elsholtz, Christian, Egyptian fractions. Nieuw Arch. Wiskd. (5) 23 (2022), no. 4, 237--245. The arXiv abstract page for the article names the Creative Commons Attribution 4.0 license for its only version, v1 (https://arxiv.org/abs/2210.04496v1, read 2026-10-02), and the term is taken from it; the held nine-page file is the journal's typeset edition, which carries no arXiv stamp and prints no notice, and the journal's issue page states no license.

This survey covers modern results on writing m/n as a sum of distinct unit fractions. It states the Erdos-Straus conjecture (Conjecture 1: for every n>=2 the fraction 4/n is a sum of three unit fractions), traces its 1950 origin, and proves Theorem 1, that the conjecture is equivalent to the statement that every prime lies in one of the congruence classes -a/c mod (4acd-1) for some a,c,d>=1, or -(4c^2 d+1)/k mod 4cd for some c,d,k>=1 with k | 4c^2 d+1, so that verification reduces to covering the primes by such classes. It reports the probabilistic heuristic that only finitely many n can fail, Bright and Loughran's proof that there is no Brauer-Manin obstruction to solutions, and the counting results: Elsholtz-Tao's sum over primes p<=N of f(p) = N(log N)^{2+o(1)} and f(p) <= p^{3/5+o(1)}, and Elsholtz-Planitzer's f(n) >= (log n)^{log 6 + o(1)} for almost all n, with the larger exp((log 6+o(1)) log n/log log n) for infinitely many n. Further sections give Vose's Theorem 2 (any m/n in (0,1) is a sum of O((log n)^{1/2}) distinct unit fractions, against Erdos's conjectured O(log log n)) and Elsholtz's Theorem 3, bounding the count E_{m,k}(N) of n<=N for which m/n is not a sum of k unit fractions by N exp(-c(log N)^{1-1/(2^{k-1}-1)}), which recovers Vaughan's bound for m=4, k=3. The whole discussion of 4/n is the reference material for Erdos problem 242, the Erdos-Straus conjecture itself.

Source: https://arxiv.org/abs/2210.04496.

The retained folder-name PDF is the typeset journal article (nine pages, printed 237--245; PDF p. n is printed p. 236+n), not the arXiv posting (arXiv:2210.04496v1, 10 October 2022, the only version listed on 2026-09-18); no Crossref record for the article was found on 2026-09-18. Read status: claims checked. Conjecture 1, Theorem 1 and Theorem 3 and the survey's sentences on Vaughan's bound and on the Elsholtz-Tao and Elsholtz-Planitzer counts were read clause by clause on the page images of pp. 239--240; the proof of Theorem 1 was read for structure; the rest of the survey was not read. Result pages: theorem_1 (the covering-congruence equivalence) and theorem_3 (the survey's restatement of Elsholtz's exceptional-set bound, which recovers Vaughan's).

Bears on. #242

Results to transcribe.

• Conjecture 1 (Erdos-Straus): For every n>=2 there are positive integers x,y,z with 4/n = 1/x + 1/y + 1/z.
• Theorem 1: The Erdos-Straus conjecture is equivalent to every prime lying in a congruence class -a/c mod (4acd-1) for some a,c,d>=1, or -(4c^2 d+1)/k mod 4cd for some c,d,k>=1 with k | 4c^2 d+1.
• Theorem 2 (Vose): Every m/n in (0,1) is a sum of O((log n)^{1/2}) distinct unit fractions.
• Theorem 3 (Elsholtz): For m>k>=3, the number of n<=N with m/n not a sum of k unit fractions is at most N exp(-c_{m,k}(log N)^{1-1/(2^{k-1}-1)}).
• Counting results (as the survey reports them on p. 240): the number f(n) of representations of 4/n as three distinct unit fractions satisfies sum_{p<=N} f(p) = N(log N)^{2+o(1)} and f(p) <= p^{3/5+o(1)} (Elsholtz-Tao); f(n) >= (log n)^{log 6 + o(1)} for almost all n, and f(n) >= exp((log 6+o(1)) log n/log log n) for infinitely many n (Elsholtz-Planitzer).