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