Problem 11721
(American Mathematical Monthly, Vol.120, August-September 2013]) Proposed by Roberto Tauraso (Italy).
Let p be a prime greater than 3, and let q be a complex number other than 1 such that qp = 1.
Evaluate
p−1
X
k=1
(1 − qk)5 (1 − q2k)3(1 − q3k)2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let a, b, d be non-negative integers such that gcd(a, p) = 1.
Let j0≡ −b/a (mod p) such that j0∈ {0, 1, . . . , p − 1}. Since
p−1
X
k=1
qk(aj+b)= −1 +
p if p | (aj + b) 0 otherwise it follows that
p−1
X
k=1
qbk
(1 − qakz)d = =
p−1
X
k=1
qbkX
j≥0
j + d − 1 d − 1
qakjzj
=X
j≥0
j + d − 1 d − 1
zj
p−1
X
k=1
qk(aj+b)
= pX
s≥0
j0+ sp + d − 1 d − 1
zj0+sp− 1 (1 − z)d
=pPd−1
s=0cszj0+sp
(1 − zp)d − 1 (1 − z)d where
cs= [zj0+sp](1 − zp)d X
s2≥0
j0+ s2p + d − 1 d − 1
zj0+s2p
= [zj0+sp] X
s1≥0
(−1)s1 d s1
zs1p X
s2≥0
j0+ s2p + d − 1 d − 1
zj0+s2p
= X
s1+s2=s
(−1)s1 d s1
j0+ s2p + d − 1 d − 1
=
s
X
k=0
(−1)s−k
d s − k
j0+ kp + d − 1 d − 1
.
Let z = 1 + w, then
p−1
X
k=1
qbk
(1 − qak)d = lim
w→0
pPd−1
s=0cs(1 + w)j0+sp−1−(1+w)p
−w
d (1 − (1 + w)p)d
= lim
w→0
pPd−1
s=0cs j0+sp
d wd+ o(wd) − (−1)dPd
s=0(−1)s ds sp
2dwd+ o(wd) (−pw + o(w))d
= 1
pd (−1)dp
d−1
X
s=0
cs
j0+ sp d
−
d
X
s=0
(−1)sd s
sp 2d
! .
Hence
p−1
X
k=1
qbk (1 − q2k)3 =
−(p − 1)(p − 3)/8 if b = 0 (p − 1)(p + 1)/24 if b = 1, 4
−(p − 1)(p + 1)/24 if b = 2, 5
0 if b = 3
,
and
p−1
X
k=1
qbk (1 − q3k)2 =
−(p − 1)(p − 5)/12 if b = 0
(p + 5)(p − 1)/36 if b = 1, 5 and p ≡ 1 (mod 3) (p − 5)(p + 1)/36 if b = 1, 5 and p ≡ −1 (mod 3)
(p − 1)2/36 if b = 2, 4 and p ≡ 1 (mod 3) (p + 1)2/36 if b = 2, 4 and p ≡ −1 (mod 3)
−(p − 1)(p + 1)/12 if b = 3
.
Therefore
p−1
X
k=1
(1 − qk)5
(1 − q2k)3(1 − q3k)2 =
p−1
X
k=1
4 − 8qk+ q2k+ 5q3k− q4k− q5k (1 − q2k)3
+
p−1
X
k=1
−3 + 3qk+ 2q3k− q4k− q5k (1 − q3k)2
=
−(55p − 1)(p − 1)/72 if p ≡ 1 (mod 3)
−(11p − 1)(5p − 1)/72 if p ≡ −1 (mod 3) .
Note that the result is always a negative integer.
