Problem 11602
(American Mathematical Monthly, Vol.118, November 2011) Proposed by Roberto Tauraso (Italy).
Let p be a prime. Let Fn denote the nth Fibonacci number. Show that X
0<i<j<k<p
Fi
ijk ≡ 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 p > 3 be prime (if p = 2, 3 the above sum is empty and it gives 0). For k = 1, . . . , p − 1
p k
= (−1)k−1p k
k−1
Y
j=1
1 −p
j
≡ (−1)k−1p
k 1 − pHk−1(1) + p2Hk−1(1, 1)
(mod p4)
2p k
= (−1)k−12p k
k−1
Y
j=1
1 − 2p
j
≡ (−1)k−12p
k 1 − 2pHk−1(1) + 4p2Hk−1(1, 1)
(mod p4)
p + k − 1 k
= p k
k−1
Y
j=1
1 + p
j
≡ p
k 1 + pHk−1(1) + p2Hk−1(1, 1)
(mod p4)
where
Hk−1(1) = X
0<i<k
1
i , Hk−1(1, 1) = X
0<i<j<k
1 ij. Let us consider the following known identities: for any integer n ≥ 0
(1)
n
X
k=0
n k
Fn−k(−1)n−k= −Fn, (2)
n
X
k=0
2n k
Fn−k(−1)n−k= −
n
X
k=0
n + k − 1 k
Fn−k.
Letting n = p, then (1) and (2) become (note that F0= 0 and p is odd)
(1) pS1− p2S2+ p3S3≡ 0 (mod p4) , (2) 2pS1− 4p2S2+ 8p3S3≡ −pS1− p2S2− p3S3 (mod p4) where
S1= X
0<i<p
Fp−i
i , S2= X
0<i<j<p
Fp−j
ij , S3= X
0<i<j<k<p
Fp−k ijk .
By subtracting 3 times (1) from (2) we obtain 6p3S3≡ 0 (mod p4), that is S3≡ 0 (mod p). Finally 0 ≡ S3= X
0<i<j<k<p
Fp−k
ijk = X
0<i<j<k<p
Fi
(p − k)(p − j)(p − i) ≡ − X
0<i<j<k<p
Fi
ijk (mod p).
