Problems & Puzzles: Puzzles

Puzzle 9.- p1 + p2 + … pk = n m , m => 2 

Let’s define S(Pk) = p1 + p2 + … + pk, namely "the sum of the first k primes"

G.L. Honaker, Jr. observes that the sum of the first 9 consecutive prime numbers = 10 2 , and asks "Can k produce other perfect squares, cubes, etc. ?"

Judson McCranie has found that S(k) is a perfect square not only for k = 9 but also for k=2474, 6694, 7785 and for … 709838… the current record for perfect squares.

Here is the expression of this record :

S(P709838) =2 + 3 + 5 + 7 + ... + 10729219 = 1916357^2

Honaker reports that no S(k) perfect cube has been found.
 

Would you like to beat the current record or to find another perfect power?

The current record solution for this old puzzle has been gotten by Giovanni Resta (May 2003):

I found the next solution to puzzle 9.That is: the sum of the primes from 2 to 3531577135439 (the 126789311423-th prime) is equal to 219704732167875184222756 which is the square of 468726713734. I double checked, so I hope this is correct.

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Zak Seidov wrote on May 5, 2018:

I try the variation of Puzzle 9 for ODD Primes
 and here are my current results
 where k is index of the last prime p and n^m = p2+p3+...pk

 [k, n^m, p]:

 [3, 8=2^3, 5]
 [24, 961=31^2, 89]
 [28, 1369=37^2, 107]
 [32, 1849=43^2, 131]
 [46, 4225=65^2, 199]
 [296, 263169=513^2, 1949]
 [5327, 130919364=11442^2, 52081]
 [12227, 758451600=27540^2, 130729]
[55560480, 29682949232484409=172287403^2, 1097877707]

 All n^m are  square except for the first.

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