Let a and b be integers and let p be a prime of the form p=2k+1. Show that if p does not divide a and a is congruent to b^2 (mod p), then a^k is congruent to 1 (mod p)

We are tasked with proving that if \( p \) is a prime of the form \( p = 2k + 1 \), \( p \) does not divide \( a \), and \( a \equiv b^2 \pmod{p} \), then \( a^k \equiv 1 \pmod{p} \). 

### Step 1: Prime form and integer relations

First, observe that \( p = 2k + 1 \) implies that \( p \) is an odd prime since \( 2k + 1 \) is always odd for any integer \( k \). This means the number of nonzero elements in the group \( \mathbb{Z}_p^* \), the multiplicative group modulo \( p \), is \( p - 1 = 2k \).

### Step 2: Setup modular conditions
We are given:
- \( a \equiv b^2 \pmod{p} \),
- \( p \) does not divide \( a \), i.e., \( \gcd(a, p) = 1 \).

From \( a \equiv b^2 \pmod{p} \), we know that there exists some \( b \in \mathbb{Z}_p \) such that:
\[
a = b^2 \pmod{p}.
\]
Thus, \( a \) is a quadratic residue modulo \( p \).

### Step 3: Euler's Criterion

Since \( p \) is an odd prime, we can apply **Euler's Criterion** for quadratic residues. Euler's Criterion states that for an integer \( b \) such that \( p \) does not divide \( b \), we have:
\[
b^{p-1} \equiv 1 \pmod{p}.
\]
Using the fact that \( p - 1 = 2k \), we get:
\[
b^{2k} \equiv 1 \pmod{p}.
\]
Thus, any element raised to the power of \( 2k \) (i.e., \( p-1 \)) is congruent to 1 modulo \( p \).

### Step 4: Showing \( a^k \equiv 1 \pmod{p} \)

Since \( a \equiv b^2 \p

Let a and b be integers and let p be a prime of the form p=2k+1. Show that if p does not divide a and a is congruent to b^2 (mod p), then a^k is congruent to 1 (mod p).

Explore a detailed proof showing that for a prime p of the form p = 2k + 1, if p does not divide a and a ≡ b^2 (mod p), then a^k ≡ 1 (mod p). This problem involves number theory, Euler's Criterion, and properties of quadratic residues. Suitable for undergraduate students.

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