<think>Okay, let's try to tackle this problem. The question is about a prime ( p \equiv 1 \pmod{4} ), and we need to compute the sum ( \sum_{n \in A_p} n ), where ( A_p = { a \in [0, p] \mid (a+1)^2 \mod p < a^2 \mod p } ). Wait, but the original problem says ( [[0,p]] ), which I think is a typo and should be ( [0, p-1] ) since we're dealing with remainders modulo ( p ). Because when...

Write s(a) for the least nonnegative residue of a^2 modulo p, i.e. s(a) = {a^2}p, for a = 0, 1, ..., p−1. Define the indicator δ(a) = 1 if s(a+1) < s(a), and δ(a) = 0 otherwise. Then A_p = {a : δ(a) = 1} and the sum in question is S_p = ∑{a∈A_p} a = ∑_{a=0}^{p−1} a δ(a).

A convenient expression for δ(a) is via floors: δ(a) = floor((s(a) + 2a + 1)/p) − floor((2a + 1)/p). This is checked case by ...

An elegant solution to this problem can be constructed by building on several results in number theory, particularly those related to character sums. The conjecture is indeed true.

The Conjecture

Let be a prime such that . Let be the remainder of on division by . The set in question is $A_p = { n \in {0, 1, \dots, p-1} \mid r_{n+1} < r_...

First, recall the setup: ( p ) is a prime with ( p \equiv 1 \pmod{4} ). The set ( A_p ) is defined as ( A_p = { a \in [0, p] \mid {(a+1)^2}_p < {a^2}_p ...

We write every square in the form

    a² = p q(a) + r(a)             (0 ≤ r(a) ≤ p–1).

Here q(a)=⌊a²/p⌋ and r(a)=a² (mod p).
Put

    Δ(a)=⌊(a+1)²/p⌋ – ⌊a²/p⌋ = ⌊(r(a)+2a+1)/p⌋ .

( a+1)² mod p is smaller than a² mod p  ⇔ Δ(a) ≥ 1.
Hence

    A p  = { a : 0≤a≤p–1 , Δ(a) ≥ 1 }, 
    1 A p (a) = min{1,Δ(a)} .                       (1)

(1) A general identity

Add and subtra...

This is an interesting number-theoretic problem involving quadratic residues and a specific sum over a set defined by a peculiar condition.

Let me analyze the problem:

Given:

• is a prime with
• denotes the remainder of upon division by

Conjecture:

**Analy...

Below is a self‐contained proof of the conjecture

 for every prime p≡1 (mod 4) one has
  ∑_{0≤n≤p, {(n+1)^2}_p<{n^2}_p} n
  = 7 (p^2−1)/24.

In fact one shows two things:

1. A “master‐formula” valid for all odd primes p≥3, (⋆) ∑{n∈A_p}n = ⎛7p^2−12p+5⎞/24 + (1/p)·∑{k=1}^{p−1}(k^2 mod p), where A_p = { 0≤n≤p:\ {(n+1)^2}_p<{n^2}_p }, and by “k^2 mod p” we mean t...

Analyzing the Mysterious Sum where

I'll analyze this intriguing number theory conjecture about the sum over the set .

Understanding the Problem

First, let's clarify what the problem is asking:

• We have a prime
• For any integer , denotes the remainder when is divided by
• $A_p = {a \in [[0,p]] \mid (a...

Let

 p ≥ 3 be an odd prime,
 n ↦ n₍ₚ₎ the least non–negative residue of n modulo p,
 Aₚ = { 0 ≤ a ≤ p-1 : (a+1)²₍ₚ₎ < a²₍ₚ₎ }, and
 S(p) = Σ_{a ∈ Aₚ} a .

The conjecture asserts that, when p ≡ 1 (mod 4),

 S(p) = 7(p² – 1)/24. (1)

We prove a slightly more general identity which immediately gives (1) and also explains what happens for p ≡ 3 (mod 4).

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