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Topic: A conjecture for sets of four primes
Replies: 12   Last Post: Oct 14, 2013 8:11 PM

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gnasher729

Posts: 418
Registered: 10/7/06
Re: A conjecture for sets of four primes
Posted: Oct 14, 2013 8:11 PM
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We look for four primes adding up to an even number, with the sum of squares adding up to a square.

Tried to see if a search can be restricted:

Since the sum is even, the sum of squares is even, and must be the square of an even number.

Even numbers are of the form 12k +/- m, where m = 0, 2, 4 or 6.
The square is 144k^2 +/- 24km + m^2 = 24n + 0, 4, 16 or 36 for some n = 24n + 0, 4, 12 or 16 for some n.

Primes are 2, 3, or 6k +/- 1. Squares of primes are 4, 9, and numbers of the form 24n + 1 for some n.
Sums of 4 squares of primes with an even sum are:

4 + 4 + 4 + 4 = 16
4 + 4 + 9 + 9 = 26
4 + 4 + 9 + (24n + 1) = 24n + 18
4 + 4 + (24n + 1) + (24m + 1) = 24n + 10
9 + 9 + 9 + 9 = 36
9 + 9 + 9 + (24n + 1) = 24n + 4
9 + 9 + (24n + 1) + (24m + 1) = 24n + 20
9 + (24n + 1) + (24m + 1) + (24k + 1) = 24n + 12
(24n + 1) + (24m + 1) + (24k + 1) + (24j + 1) = 24n + 4

4 + 4 + 4 + 4 and 9 + 9 + 9 + 9 are squares.
9 + 9 + 9 + (24n + 1) = 24n + 4 is a possible square (12k +/- 2)^2
9 + (24n + 1) + (24m + 1) + (24k + 1) = 24n + 12 is a possible square (12k + 6)^2
(24n + 1) + (24m + 1) + (24k + 1) + (24j + 1) = 24n + 4 is a possible square (12k +/- 2)^2.

9 + 9 + 9 + p^2 cannot be a square if p > 13 since it is less than (p + 1)^2. It is a square if and only if p = 13. So the possible 4-tuples of primes adding up to an even number with sum of squares being a square are: (2, 2, 2, 2), (3, 3, 3, 3), (3, 3, 3, 13), and (p, q, r, s) where p <= q <= r <= s, p >= 3, q >= 5.

I may check how many numbers have a solution where the first prime is 3; probably most.



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