Problems & Puzzles:
of consecutive squared primes a square
Here we ask for solutions to: S(pi2)=
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:
+ ... +1732
Find solutions for:
p1=2 or for p1=3
A = prime (no matter what p1 is)
Just before turning the page, find one solution to S(pi2)
both sums run over two distinct but contiguous sets of consecutive primes.
+ 6472) = (9412 +
is one almost-solution where unfortunately the two sets are not contiguous...
that is to say this not a solution for the c) question.
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.).
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.