Problems & Puzzles: Puzzles

Puzzle 128.  Sum of consecutive squared primes a square

Here we ask for solutions to:  S(p_i²)= 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(p_i²) =  S(q_i²) 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.

______
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.

***

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.
 

***