Problem 11728
(American Mathematical Monthly, Vol.120, October 2013) Proposed by Walter Blumberg (USA).
Letp be a prime congruent to 7 modulo 8. Prove that
p
X
k=1
k2+ k p
= 2p2+ 3p + 7
6 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let rn(m) be the remainder of the division of m by n then, since m = n⌊m/n⌋ + rn(m), it follows that
p
X
k=1
k2+ k p
= 1 p
p
X
k=1
(k2+ k) −1 p
p
X
k=1
rp(k2+ k) = p2+ 3p + 2
3 −1
p
p−1
X
k=1
rp(k2+ k)
Note that rp(k2+ k) = j for some j ∈ {1, . . . , p − 1} iff k2+ k = j (mod p), that is iff ∆ = 1 + 4j is a square modulo p iff
4j+1 p
= 1 where
· p
is the Legendre symbol. Therefore the number of k ∈ {1, . . . , p − 1} such that rp(k2+ k) = j is4j+1
p
+ 1
(it gives 0 or 2) and
p−1
X
k=1
rp(k2+ k) =
p−1
X
j=1
j 4j + 1 p
+ 1
=
p−1
X
j=1
j 4j + 1 p
+p2− p 2 .
Hence,
p
X
k=1
k2+ k p
=p2+ 3p + 2
3 −1
p
p−1
X
j=1
j 4j + 1 p
−p − 1
2 =2p2+ 3p + 7
6 −1
p
p−1
X
j=0
j 4j + 1 p
,
so suffices to show that
S1=
p−1
X
j=0
j 4j + 1 p
= 0.
Let
A =
p−1
X
j=0
j j p
, B =
p−1
X
j=0
j p
,
Ui=
p−1
X
j=0
j 2j + i p
, Vi=
p−1
X
j=0
2j + i p
for i = 0, 1,
Si=
p−1
X
j=0
j 4j + i p
, Ti=
p−1
X
j=0
4j + i p
for i = 0, 1, 2, 3.
SincePp−1 j=0
aj+b p
= 0 when p be a prime not dividing a, it follows that B = Vi= Ti= 0 because p does not divide 2 and 4. Moreover p ≡ 7 (mod 8) yields
−1 p
= (−1)(p−1)/2= −1 and 2 p
= (−1)⌊(p+1)/4⌋= 1.
Now, U0= S0= A and
2p−1
X
r=0
r r p
=
1
X
i=0 p−1
X
j=0
(2j + i) 2j + i p
= 2U0+ 2U1+ V1= 2A + 2U1,
2p−1
X
r=0
r r p
=
1
X
i=0 p−1
X
r=0
(ip + r) ip + r p
= A + pB + A = 2A,
which imply that S2= U1= 0. Moreover
4p−1
X
r=0
r r p
=
3
X
i=0 p−1
X
j=0
(4j + i) 4j + i p
= 4S0+ 4S1+ 4S2+ 4S3+ T1+ T2+ T3= 4A + 4S1+ 4S3,
4p−1
X
r=0
r r p
=
3
X
i=0 p−1
X
r=0
(ip + r) ip + r p
= A + pB + A + 2pB + A + 3pB + A = 4A,
which imply that S1+ S3= 0. Finally
S1=
p−1
X
j=1
j 4j + 1 p
=
p−1
X
j=1
(p − j) 4(p − j) + 1 p
= −p
p−1
X
j=1
4j − 1 p
+
p−1
X
j=1
j 4j − 1 p
= −(p − 1)
p−2
X
j=0
4j + 3 p
+
p−2
X
j=0
j 4j + 3 p
= (p − 1) −1 p
− (p − 1)T3− (p − 1) −1 p
+ S3= S3
and we obtain that S1= S3= 0.
