Problem 10454
(American Mathematical Monthly, Vol.102, May 1995) Proposed by H. Tamvakis (USA).
We say that a natural number n is amenable if there exist integers a1, a2, . . . , an such that a1+ a2+ . . . + an= a1a2· · · an= n.
Find all amenable numbers.
Solution proposed by Roberto Tauraso, Scuola Normale Superiore, piazza dei Cavalieri, 56100 Pisa, Italy.
Theorem. LetA be the set of all amenable numbers then
A = {n ∈ N∗ : [n]4= [1]4or[n]4= [0]4} \ {4}.
Note thatA is closed under multiplication.
Proof. Let n ∈ N.
(a) If [n]4= 1 then n ∈ A because it suffices to choose
a1= n and
a1+i= −1 for i = 1, . . . ,n−12 a1+n−1
2 +i= 1 for i = 1, . . . ,n−12 . (b) If [n]4= 0 and n > 4 then n ∈ A. If n4 is even then choose
a1=n
2, a2= 2 and
a2+i= −1 for i = 1, . . . ,n4 a2+n4+i= 1 for i = 1, . . . ,3n4 − 2 . Whereas, if n4 is odd then choose
a1=n
2, a2= −2 and
a2+i= −1 for i = 1, . . . ,n4 − 2 an4+i = 1 for i = 1, . . . ,3n4 . (c) 4 6∈ A: if 4 = a1a2a3a4then it is easy to verify that a1+ a2+ a3+ a46= 4.
Now, we assume that n ∈ A. Define the sets
I+= {1 ≤ i ≤ n : ai >0} and I− = {1 ≤ i ≤ n : ai<0}
and let u, v be their cardinalities.
Since the equation of the statement holds, then v is even and X
i∈I+
ai+ X
i∈I−
ai= u + v = n.
Hence
X
i∈I+
ai− u + X
i∈I−
ai+ v = 2v and this means that
[X
i∈I+
(ai− 1) + X
i∈I−
(ai+ 1)]4= [0]4. Now, we use the above equation to complete the proof.
(d) If [n]4 = 2 then n 6∈ A: let n = a1a2· · · an then one and only one of the factors ai is even, therefore
[X
i∈I+
(ai− 1) + X
i∈I−
(ai+ 1)]4∈ {[1]4,[3]4} contradicting the previous equation.
(e) If [n]4= 3 then n 6∈ A: let n = a1a2· · · anthen all the factors are odd and for an odd number of them, [ai]4= 3, thus
[X
i∈I+
(ai− 1) + X
i∈I−
(ai+ 1)]4= [2]4
contradicting the previous equation.
