
Re: A conjecture for sets of four primes
Posted:
Oct 14, 2013 8:11 PM


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 4tuples 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.

