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# We show that such partition exists if and only if n is a multiple of 8 with n ≥ 16

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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. 

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