Erdős–Straus conjecture
Precise statement and status
Status: open. For every integer n >= 2 there exist positive integers x, y, z such that
4/n = 1/x + 1/y + 1/z.
Denominators need not be distinct. Since the summands are symmetric, searches may assume x <= y <= z without loss of generality.
What is known
- The conjecture concerns three-term Egyptian fraction representations. MathWorld’s overview supplies references and reports verification through 10^14; that is a sourced historical bound, not a claim about today’s record, and has not been rerun here.
- Elementary reduction (derivation included here): if (x,y,z) works for m, then (kx,ky,kz) works for km, by dividing the identity by k. Every n >= 2 has a prime divisor, so proving the statement for all primes would suffice.
- Even n work via (x,y,z) = (n/2,n,n): the right side is 2/n + 1/n + 1/n. This is a direct identity, not a solution for all n.
- Our first computation finds and checks one solution for each n from 2 through 100.
Why it is hard
Many algebraic identities cover particular residue classes, but covering every remaining prime is another matter. A search with an arbitrary denominator cutoff can miss solutions; large computational coverage still leaves infinitely many inputs.
Small entry points
- Review the seed’s derived finite search bounds and integer identity checks.
- Give a proved identity for a specific residue class, spelling out positivity and integrality conditions.
- Apply the scaling reduction to a few composite inputs and compare with the seed’s first-found solutions.
Contributions
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