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This post posed an interesting question; its answer is negative. Now, we take the sum of squares into consideration. Let $a_1,a_2,a_3...a_k$ be $k$ positive integers such that any two of them are different.

Caculate: $$A=\displaystyle \sum_{i=1}^{k} a_i$$ $$B=\displaystyle \prod_{i=1}^{k} a_i$$ $$C=\displaystyle \sum_{i=1}^{k} a_i^{2}$$

$A,B$, $C$ and $k$ are known to us. Can we uniquely identify these $k$ positive integers? I think the answer is probably no, but I can't find a counterexample.

  • 1
    Surely if $k=3$ we can identify those numbers as the roots of the polynomial $$x^3-Ax+((A^2-C)/2)x-B$$ Hence, any counterexample needs $k \ge 4$. – 2017-02-02
  • 1
    very unlikely if $k \ge 4$ – 2017-02-02
  • 0
    For each $k$ is necessary to solve separately the equation. I don't understand what there can be problems. First, it is necessary to solve 1 and 3 the equation then to find the parameters satisfying 2. – 2017-02-02

1 Answers 1

5

k=4, $a_i$ pairwise different

This seems to be the minimal solution in terms of A (i.e. there seems to be no solution with $a_1+a_2+a_3+a_4 < 42$):

$a_i$ = (4, 5, 15, 18) and $a_i$ = (3, 9, 10, 20) both yield (A,B,C) = (42, 5400, 590)

Next there are two solutions for A=45:

(45, 8640, 641): (5, 6, 16, 18), (4, 9, 12, 20)
(45, 4224, 765): (3, 4, 16, 22), (2, 8, 11, 24)

And many more:

(48, 7200, 842): (4, 5, 15, 24), (3, 8, 12, 25)
(48, 12600, 710): (6, 7, 15, 20), (5, 10, 12, 21)
(52, 3300, 1122): (2, 3, 22, 25), (1, 10, 11, 30)
(54, 15400, 958): (5, 7, 20, 22), (4, 11, 14, 25)
(55, 6720, 1209): (3, 4, 20, 28), (2, 7, 16, 30)
(55, 20160, 941): (6, 8, 20, 21), (5, 12, 14, 24)
(56, 8580, 1194): (3, 5, 22, 26), (2, 11, 13, 30)
(56, 12600, 1118): (4, 6, 21, 25), (3, 10, 15, 28)
(60, 4104, 1670): (2, 3, 19, 36), (1, 9, 12, 38)
(60, 7560, 1574): (3, 4, 18, 35), (2, 7, 15, 36)
(60, 9900, 1418): (3, 5, 22, 30), (2, 10, 15, 33)
(60, 13464, 1430): (4, 6, 17, 33), (3, 11, 12, 34)
(60, 22680, 1214): (6, 7, 20, 27), (5, 9, 18, 28)
(60, 29484, 1130): (7, 9, 18, 26), (6, 13, 14, 27)
(60, 33264, 1070): (8, 9, 21, 22), (7, 11, 18, 24)
(60, 41580, 986): (10, 11, 18, 21), (9, 14, 15, 22)
...

k=4, some of the $a_i$ may be equal

Without the restriction that all $a_i$ must be pairwise different, there are additional smaller solutions:

(30, 1260, 314): (3, 3, 10, 14), (2, 6, 7, 15)
(35, 2880, 401): (4, 4, 12, 15), (3, 6, 10, 16)
(36, 924, 570): (2, 2, 11, 21), (1, 6, 7, 22)
(36, 4200, 390): (5, 5, 12, 14), (4, 7, 10, 15)
(40, 2520, 614): (3, 3, 14, 20), (2, 5, 12, 21)
(42, 1300, 802): (2, 2, 13, 25), (1, 5, 10, 26)

k=5, $a_i$ pairwise different

(43, 5400, 591): (1, 4, 5, 15, 18), (1, 3, 9, 10, 20)
(44, 10800, 594): (2, 4, 5, 15, 18), (2, 3, 9, 10, 20)
(44, 15120, 570): (3, 4, 5, 14, 18), (2, 6, 7, 9, 20) *
(46, 4224, 766): (1, 3, 4, 16, 22), (1, 2, 8, 11, 24)
(46, 8064, 726): (2, 3, 4, 16, 21), (1, 6, 7, 8, 24) *
(46, 8640, 642): (1, 5, 6, 16, 18), (1, 4, 9, 12, 20)
(47, 17280, 645): (2, 5, 6, 16, 18), (2, 4, 9, 12, 20)
...

Interestingly the solutions start around the same values for both A and max($a_i$):

k=5, some of the $a_i$ may be equal

(30, 3600, 234): (3, 3, 4, 10, 10), (2, 5, 5, 6, 12) *
(31, 1260, 315): (1, 3, 3, 10, 14), (1, 2, 6, 7, 15)
(32, 2520, 318): (2, 3, 3, 10, 14), (2, 2, 6, 7, 15)
(33, 3780, 323): (3, 3, 3, 10, 14), (2, 3, 6, 7, 15)
(34, 5040, 330): (3, 3, 4, 10, 14), (2, 4, 6, 7, 15)
(35, 6300, 339): (3, 3, 5, 10, 14), (2, 5, 6, 7, 15)
(36, 2880, 402): (1, 4, 4, 12, 15), (1, 3, 6, 10, 16)
(36, 7560, 350): (3, 3, 6, 10, 14), (2, 6, 6, 7, 15)
(37, 924, 571): (1, 2, 2, 11, 21), (1, 1, 6, 7, 22)
...

[solutions for k=5 marked with '*' cannot be constructed from a solution for k=4 in a trivial way (by adding the same value to both tuples)]

  • 0
    If I may ask, how did you find this? Do you have more? – 2017-02-02
  • 0
    I checked it and it proves to be completely valid! Thank you! – 2017-02-02
  • 0
    @J.Doe guessing? no: a few lines of Python: `s = {}` `for N in range(4,50):` `for a1 in range(1,N-2):` `for a2 in range(a1+1,N-1):` `for a3 in range(a2+1,N):` `ai=(a1,a2,a3,N)` `ABC=(a1+a2+a3+N,a1*a2*a3*N,a1*a1+a2*a2+a3*a3+N*N)` `if ABC in s: print(...)` `else: s[ABC]=ai` – 2017-02-03