Problems & Puzzles: Puzzles

Puzzle 904. Follow-up to Puzzles 902 & 903 Emmanuel Vantieghem sent the following follow-up to Puzzles 902 & 903: Given P&Q such that Q=P could these be combined by the sumproduct(P, Q) operation giving a power b^n, n>1 of b>1? Examples by Emmanuel: P = (41, 47, ... , 173)  (28 primes) : P x P = 586^2  P = (157, 163, ... , 1097)  (148 primes) : P x P = 8174^2 P = (547, 557, ... , 971)  (64 primes) : P x P = 6088^2 P =(50891, 50893, ... , 51199} (28 primes) : P x P = 270106^2  Emmanuel wrote "I could not find third or higher powers". By my way I [CR] found one cubic solution: P = (477973,...478417) (35 primes) : PxP=20003^3 Q. Find more vectors P such that sumproduct(P, P)=b^n for some b>2 & n>2.

---

Contribution came from Dmitry Kamenetsky

***

Dmitry wrote (Dec. 18, 2107):

For n>2 I couldn't find any more solutions for b^n<9*10^18. For n=2, there is a sequence in the OEIS: oeis.org/A270424

***

Jeff Heleen wrote on Dec 29, 2017:

I have searched starting primes up to 23,000,000 for b^n, with 3 <= n <= 15 and
for vectors P & Q and P=Q with lengths k <= 10000. No solutions were found other than the one by [CR]

***

---