Problems & Puzzles: Puzzles

Puzzle 128.  Sum of consecutive squared primes a square

Here we ask for solutions to:  S(pi2)= A2 where the sum runs over consecutive primes and A is integer.

If we do not restrict what the first prime (p1) is, the least solution (regarding the A value) is:

412 + ... +1732 = 5862

Questions: Find solutions for:

a)       p1=2 or for p1=3

b)       A = prime (no matter what p1 is)

c) Just before turning the page, find one solution to S(pi2)S(qi2) where both sums run over two distinct but contiguous sets of consecutive primes.
(1312 +… + 6472) = (9412 +…+ 10332) is one almost-solution where unfortunately the two sets are not contiguous... that is to say this not a solution for the c) question.

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Encouraging note: maybe you'll find interesting to know that S(n2)= N2, n = 1 to k has one and only one conspicuous solution (Cfr. p. 77 Excursions in number theory, C. S. Ogilvy & J. T. Anderson, Dover Publications, Inc.).

Clarifying note: None of my two examples are solutions to the asked questions. They are mere illustrations of the numerical expressions.

Giovanni Resta wrote (Nov. 2004):

Searched without success for initial point equal to 2 or 3 up to 1,693,930,336,951.
Using different initial points (p_1 < 6,461,335,109) and sequence length less than 1000, I found only one solution, namely

2489647^2 = 355363^2 + (other 47 terms) + 355951^2

It is nice since we have a sum of 49=7^2 quadratic terms.

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Jim Fougeron wrote (Apr., 19, 2007):

Here are a couple more solutions to question 'b' :

27847739^2 == 290209^2 + (other 6911 terms) + 378193^2 (validated A^2 and summation are both 775496567412121)

74930959^2 == 654889^2 + (other 10607 terms) + 797869^2 (validated A^2 and summation are both 5614648616659681)

Note, 10609==103^2 (Giovanni was also a perfect square number of factors), however, 6913 is 31*223, so the square number of factors does not hold for all solutions, but it does seem interesting.

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