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# Let c(n, j) =X S Y k∈S k where the sum is over all j-element subsets S of the set {1

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Problem 11259

(American Mathematical Monthly, Vol.113, December 2006) Proposed by N. Abe (Japan).

For integers n greater that 2, let

f(n) =

n−2

X

j=1

2jX

S

Y

k∈S

k,

where the sum is over all j-element subsets S of the set{1, . . . , n − 1}.

Show that4(2n − 1)! + (f (n))2 is never the square of an integer.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let

c(n, j) =X

S

Y

k∈S

k

where the sum is over all j-element subsets S of the set {1, . . . , n − 1}. We prove that this number is actually a Stirling number of the first kind

c(n, j) =

 n n− j

 .

In fact, c(n, 0) =n

n = 1 and c(n, n − 1) = n1 = (n − 1)! for all n ≥ 1. Moreover for n, j ≥ 2

c(n, j) = X

(n−1)6∈S

Y

k∈S

k+ X

(n−1)∈S

Y

k∈S

k

= c(n − 1, j) + (n − 1) · c(n − 1, j − 1) =

 n− 1 n− 1 − j



+ (n − 1)n − 1 n− j



=

 n n− j

 . Hence for n > 2 we have that

f(n) =

n−2

X

j=1

2jc(n, j) =

n−2

X

j=1

2j

 n n− j



= 2n

n−2

X

j=1

1 2n−j

 n n− j



= 2n

n−1

X

k=2

1 2k

n k



= 2n  1 2

n

− 1 21

n 1



− 1 2n

n n

!

= (2n − 1)!! − 2n−1(n − 1)! − 1 = (2n − 1)!! − (2n − 2)!! − 1 where we used the identity

xn=

n

X

k=1

n k

 xk.

Assume that for n > 2 there is a number q such that 4(2n − 1)! + (f (n))2= q2. Letting d = (2n − 1)!! and p = (2n − 2)!! we have that

4(2n − 1)! = 4pd = (d + p)2− (d − p)2= q2− (f (n))2= q2− (d − p − 1)2 that is

0 < 2d − 2p − 1 = (d − p)2− (d − p − 1)2= (d + p)2− q2≥ (d + p)2− (d + p − 1)2= 2d + 2p − 1

which implies that 4p ≤ 0 and we get a contradiction. 

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