Problems & Puzzles: Puzzles

Puzzle 613. Curio sum of primes.

Robert D. Mohr sent the following nice curio & puzzle:

Consider the following seven groups of four distinct primes:

 Set p1 p2 p3 p4 1 5 7 23 29 2 37 89 97 101 3 113 229 277 281 4 179 197 317 331 5 281 331 337 347 6 359 389 421 431 7 3361 3467 3779 3793

For each set, both a) the sum of the primes & b) the sum of the squared primes, is a square.

Example:

5+7+23+29 = 8x8
5*5+7*7+23*23+29*29 = 38x38

Questions:

Q1 Are there infinite sets like these above? (send your largest p4-example)

Computing R=(p4-p1)/p1, you may verify that the R value for each set is decreasing (4.8, 1.73, 1.49, ..., 0.13)

Q2. Send a set with your smallest R value.

Q3 Are sets of four primes giving a square for the sum of primes, for the sum of squared primes & for the sum of cubed primes?

Q4 Are sets of k<>4 primes such that both a) the sum of the primes & b) the sum of the squared primes, is a square?

Contribution came from Emmanuel Vantieghem & Hakan Summakoğlu

***

Emmanuel wrote:

Q1 & Q2: My largest  p4 example is also my smallest R-value example :

{951389, 952619, 953539, 968389}  with R-value :  0.0178686.

Q3: This a set of primes with sum, sum of squares and sum of cubes square :

{17, 397, 2081, 4229}.

Q4 : since a square is always 1 or 4 modulo 4, k must be either 2, 4m or 4m+1.

For k = 2  gives no solutions (easy to see).

For k = 4m+1, we must have p1 = 2.

k=5: {2, 3, 5, 11, 79}.

k=9: {2, 5, 7, 11, 13, 17, 23, 97, 401}.

For k = 8, all primes should be odd.  I found many solutions, the 'smallest' one is :

{3, 5, 7, 11, 13, 19, 29, 109}.

***

Hakan wrote:

Q1: I wrote only the first 20 sets.

1   3 5 19 37
2   7 13 53 503
3   11 59 149 181
...
20   227 281 331 457
….

I guess  there are infinite sets.

Q2:

r=(8573-8243)/ 8243=0.04003
p1=8243, p2=8501, p3=8539, p4=8573

Q4:

For k=5, Set:2 3 5 11 79
For k=6, Set:2 5 7 19 29 59
For k=8, Set:3 5 7 11 19 37 53 61
For k=9, Set:2 5 7 11 13 17 47 53 101

***

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