Problem 11364
(American Mathematical Monthly, Vol.115, May 2008) Proposed by P. P. Dalyay (Hungary).
Let p be a prime greater than 3, and t the integer nearest p/6.
(a) Show that if p = 6t + 1, then
(p − 1)!
2t−1
X
j=0
(−1)j 3j + 1 +
2t−1
X
j=0
(−1)j 3j + 2
≡0 (mod p).
(b) Show that if p = 6t − 1, then
(p − 1)!
2t−1
X
j=0
(−1)j 3j + 1 +
2t−2
X
j=0
(−1)j 3j + 2
≡0 (mod p).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let ω = e2πi/3, then
p−1
X
k=1
ωkp k
= (1 + ω)p−1 − ωp= e(6t±1)πi/3−1 − e2(6t±1)πi/3= e±πi/3−1 − e±2πi/3= 0 On the other hand, if p = 6t + 1 then the rhs can be written as follows
p−1
X
k=1
ωkp k
=
2t
X
j=1
ω3j p 3j
+
2t−1
X
j=1
ω3j+1
p 3j + 1
+
2t−1
X
j=0
ω3j+2
p 3j + 2
=
2t
X
j=1
p 3j
p − 1 3j − 1
+
2t−1
X
j=0
ω p 3j + 1
p − 1 3j
+
2t−1
X
j=0
ω2 p 3j + 2
p − 1 3j + 1
= p
2t
X
j=1
1 3j
p − 1 3j − 1
+
2t−1
X
j=0
ω 3j + 1
p − 1 3j
+
2t−1
X
j=0
ω 3j + 2
p − 1 3j + 1
Therefore, by taking the imaginary part we have that
2t−1
X
j=0
1 3j + 1
p − 1 3j
−
2t−1
X
j=0
1 3j + 2
p − 1 3j + 1
= 0.
Since for 0 ≤ k ≤ p − 1 (p − 1)!p − 1
k
= (p − 1)(p − 2) · · · (k + 1) · (p − 1)(p − 2) · · · (p − k) ≡ (−1)k+1 (mod p) then the above identity gives (a).
If p = 6t − 1 then the rhs of the first equation can be written as follows
p−1
X
k=1
ωkp k
=
2t−1
X
j=1
ω3j p 3j
+
2t−1
X
j=1
ω3j+1
p 3j + 1
+
2t−2
X
j=0
ω3j+2
p 3j + 2
=
2t−1
X
j=1
p 3j
p − 1 3j − 1
+
2t−1
X
j=0
ω p 3j + 1
p − 1 3j
+
2t−2
X
j=0
ω2 p 3j + 2
p − 1 3j + 1
= p
2t
X
j=1
1 3j
p − 1 3j − 1
+
2t−1
X
j=0
ω 3j + 1
p − 1 3j
+
2t−2
X
j=0
ω 3j + 2
p − 1 3j + 1
Hence, by taking the imaginary part we have that
2t−1
X
j=0
1 3j + 1
p − 1 3j
−
2t−2
X
j=0
1 3j + 2
p − 1 3j + 1
= 0
which gives (b).
