Problem 12085
(American Mathematical Monthly, Vol.126, January 2019) Proposed by J. DeVincentis, S. Wagon, and M. Elgersma (USA).
For which positive integers n can{1, . . . , n} be partitioned into two sets A and B of the same size so that
X
k∈A
k=X
k∈B
k, X
k∈A
k2= X
k∈B
k2, and X
k∈A
k3=X
k∈B
k3?
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We show that such partition exists if and only if n is a multiple of 8 with n ≥ 16.
Let n be a positive integer such that the partition exists. Since |A| = |B|, then n has to be even.
Hence n + 1 is odd and P
k∈Ak = 12Pn
k=1k= n(n+1)4 ∈ N implies that 4 divides n. Let n = 4N then, since k ≡ k3 (mod 2),
(P
k∈Ak=12Pn
k=1k=n(n+1)4 = N (4N + 1) ≡ N (mod 2) P
k∈Ak3= 12Pn
k=1k3= n2(n+1)8 2 = 2N2(4N + 1)2≡ 0 (mod 2)
which implies that N ≡ 0 (mod 2) and therefore n has to be a multiple of 8. It can be verified that there is no such partition for n = 8. It remains to show that the partition exists when n is multiple of 8 and n ≥ 16.
For any non-negative integer x, we have a partition of {1 + x, . . . , 8 + x},
A8(x) := {1 + x, 4 + x, 6 + x, 7 + x)} and B8(x) := {2 + x, 3 + x, 5 + x, 8 + x}
and a partition of {1 + x, . . . , 12 + x}
A12(x) := {1+x, 3+x, 7+x, 8+x, 9+x, 11+x)} and B12(x) := {2+x, 4+x, 5+x, 6+x, 10+x, 12+x}
such that |A8(x)| = |B8(x)|, |A12(x)| = |B12(x)| and for j = 1, 2, X
k∈A8(x)
kj= X
k∈B8(x)
kj and X
k∈A12(x)
kj = X
k∈B12(x)
kj.
Therefore, if m is positive integer of the form 8a + 12b = 4(2a + 3b) with a, b ≥ 0, that is if m is a multiple of 4 with m ≥ 8, then we have a partition of {1, . . . , m},
A′:=
a−1
[
j=0
A8(8j) ∪
b−1
[
j=0
A12(8a + 12j) and B′ :=
a−1
[
j=0
B8(8j) ∪
b−1
[
j=0
B12(8a + 12j)
such that |A′| = |B′| and X
k∈A′
k= X
k∈B′
k, and X
k∈A′
k2= X
k∈B′
k2.
Finally we define a partition of {1, . . . , n} with n = 2m,
A:= A′∪ (m + B′) and B := B′∪ (m + A′).
Then it is easy to verify that |A| = |B|,P
k∈Ak=P
k∈Bk,P
k∈Ak2=P
k∈Bk2 and X
k∈A
k3= X
k∈A′
k3+ X
k∈B′
(m + k)3= X
k∈A′
k3+ m3|B′| + 3m2 X
k∈B′
k+ 3m X
k∈B′
k2+ X
k∈B′
k3
= X
k∈B′
k3+ m3|A′| + 3m2 X
k∈A′
k+ 3mX
k∈A′
k2+ X
k∈A′
k3=X
k∈B
k3
and we are done.
