Letp and q be distinct primes with p ≡ q (mod 4)

Problem 11529

(American Mathematical Monthly, Vol.117, October 2010) Proposed by W. Blumberg (USA).

Forn ≥ 1, let

An = 3

n

X

k=1

 k² n



− n².

Letp and q be distinct primes with p ≡ q (mod 4). Show that Apq= Ap+ Aq− 2.

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), we have that

An= 3

n

X

k=1

 k² n



− n²= 3 n

n

X

k=1

k²−3 n

n

X

k=1

rn(k²) − n²

= (n + 1)(2n + 1)

6 − n²− 1 n

n

X

k=1

rn(k²) = 3n + 1 2 −3

n

n−1X

k=0

rn(k²).

Let p and q be two distinct odd primes (by hypothesis p and q are different from 2). Note that rp(k²) = j for some j ∈ {1, . . . , p − 1} iff k²= j mod p, that is iff j is a square mod p which means that_j

p

= 1 where

p·

is the Legendre symbol. Therefore the number of k ∈ {1, . . . , p − 1} such that rp(k²) = j is

j p

+ 1 (it gives 0 or 2) and

p−1

X

k=0

rp(k²) =

p−1

X

j=0

 j p

 + 1

 j.

In a similar way, by the Chinese Remainder Theorem,

pq−1

X

k=0

rpq(k²) =

pq−1

X

j=0

 j p



+ 1  j q

 + 1

 j.

Hence

Ap= 2 −3 p

p−1

X

j=0

 j p



j, Aq = 2 −3 q

q−1

X

j=0

 j q

 j and

Apq= 2 − 3 pq

pq−1

X

j=0

 j p

  j q

 j − 3

pq

pq−1

X

j=0

 j p

 j − 3

pq

pq−1

X

j=0

 j q

 j.

SincePp−1 j=0

_j

p

= 0, it follows that

pq−1

X

j=0

 j p

 j =

q−1

X

a=0 p−1

X

r=0

 ap + r p



(ap + r) = p

q−1

X

a=0

a

p−1

X

r=0

 r p

 + q

p−1

X

r=0

 r p

 r = q

p−1

X

r=0

 r p

 r.

So, it suffices to prove that for p = q (mod 4)

pq−1

X

j=0

 j p

  j q

 j = 0.

We first notice that p = q = 1 (mod 4) or p = q = 3 (mod 4) and in both cases

 −1 p

  −1 q



= (−1)(p−1)/2+(q−1)/2= 1

which implies

pq−1

X

j=1

 j p

  j q

 j =

pq−1

X

j=1

 pq − j p

  pq − j q



(pq−j) =

pq−1

X

j=1

 −j p

  −j q



(pq−j) =

pq−1

X

j=1

 j p

  j q

 (pq−j)

that is

2

pq−1

X

j=0

 j p

  j q

 j = pq

pq−1

X

j=0

 j p

  j q

 .

Moreover

pq−1

X

j=0

 j p

  j q



=

q−1

X

a=0 p−1

X

r=0

 ap + r p

  ap + r q



=

p−1

X

r=0

 r p

q−1

X

a=0

 ap + r q



=

p−1

X

r=0

 r p

q−1

X

k=0

 k q



= 0

because p and q are distinct primes and

{ap + r (mod q) : a = 0, . . . , q − 1} = {k (mod q) : k = 0, . . . , q − 1}.