Problem 11426
(American Mathematical Monthly, Vol.116, April 2009) Proposed by M. L. Glasser (USA).
Find Γ(1/14)Γ(9/14)Γ(11/14)
Γ(3/14)Γ(5/14)Γ(13/14).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will prove the following generalization: let p be a prime such that p = 7 mod 8 then
Q(p) := Y
a∈O(2p)
Γ a
2p
(ap)
= 2P(p−1)/2k=1 (kp)
where O(s) is the set of odd positive integers less than s and
· p
is the Legendre symbol. In the particular case when p = 7 we have that
Q(7) = Γ(1/14)Γ(9/14)Γ(11/14)
Γ(3/14)Γ(5/14)Γ(13/14)= 2(17)+(27)+(37) = 2.
Since p = 7 mod 8
2p − a p
= −a p
= −1 p
a p
= (−1)(p−1)/2 a p
= − a p
then
Q(p) = Y
a∈O(p)
Γ(a/2p) Γ((2p − a)/2p)
(ap) .
Γ(x) satisfies the reflection formula and the duplication formula:
Γ(x)Γ(1 − x) = π
sin(πx) , Γ(x)Γ
x + 1
2
= Γ(2x)21−2x√ π, therefore we have that for x 6= 1, 1/2
Γ(x)
Γ(1 − x) = Γ(x)2sin(πx)
π = Γ(2x)2 Γ x + 122
sin(π(1 − 2x)) sin(π/2(1 − 2x))21−4x
= Γ(2x)2 Γ x +122
Γ x + 12 Γ 12− x Γ(2x)Γ(1 − 2x) 21−4x
= Γ(2x)Γ 12− x
Γ x +12 Γ(1 − 2x)21−4x. Hence
Q(p) = Y
a∈O(p)
2(1−2ap)(ap)
Γ
a p
Γ1 2(p−a)
p
Γ
p−a p
Γ1 2(p+a)
p
(ap)
.
Moreover, since p = 7 mod 8 and a is an odd integer non divisible by p
1 2(p ± a)
p
= 2 p
−1
p ± a p
= (−1)−(p2−1)/8 ±a p
= ± a p
.
So
Q(p) = Y
a∈O(p)
2(1−2ap)(ap)
Γ
a p
(ap) Γ
p−a p
(p−ap )
Γ1 2(p+a)
p
1 2(p+a)
p
Γ1 2(p−a)
p
1 2(p−a)
p
= Y
a∈O(p)
2(1−2ap)(ap) Qp−1
k=1
Γ
k p
(kp)
Qp−1 k=1
Γ
k p
(kp) = 2Pa∈O(p)(1−2ap)(ap).
Since
2 p
= 1 and
−1 p
= −1 then we can simplify the exponent:
X
a∈O(p)
a p
=
(p−1)/2
X
k=1
p − 2k p
= −
(p−1)/2
X
k=1
k p
,
and
X
a∈O(p)
a p
a p
=
p−1
X
k=1
k p
k p
−
(p−1)/2
X
k=1
2k p
2k p
=
p−1
X
k=(p+1)/2
k p
k p
−
(p−1)/2
X
k=1
k p
k p
=
(p−1)/2
X
k=1
p − k p
p − k p
−
(p−1)/2
X
k=1
k p
k p
= −
(p−1)/2
X
k=1
k p
.
Finally
Q(p) = 2Pa∈O(p)(1−2ap)(pa) = 2P(p−1)/2k=1 (kp).
