Let's solve the problem in **2 steps**, explaining each step: --- ### **Step 1: State Euler's Criterion and Apply to -1** **Euler's Criterion:** Let \( p \) be an odd prime and \( a \) an integer not divisible by \( p \). Then \[ a \text{ is a quadratic residue mod } p \iff a^{\frac{p-1}{2}} \equiv 1 \pmod{p} \] and \[ a \text{ is a non-residue mod } p \iff a^{\frac{p-1}{2}} \equiv -1 \pmod{p} \] Apply this to \( a = -1 \): \[ (-1)^{\frac{p-1}{2}} \equiv \begin{cases} 1 \pmod{p} & \text{if } -1 \text{ is a quadratic residue mod } p\\ -1 \pmod{p} & \text{if } -1 \text{ is a non-residue mod } p \end{cases} \] **Explanation:** We use Euler's Criterion with \( a = -1 \) to relate the residue of \(-1\) to the parity of \( \frac{p-1}{2} \). --- ### **Step 2: Analyze the Exponent \(\frac{p-1}{2}\) and the Result** \[ (-1)^{\frac{p-1}{2}} = \begin{cases} 1 & \text{if } \frac{p-1}{2} \text{ is even}\\ -1 & \text{if } \frac{p-1}{2} \text{ is odd} \end{cases} \] - \( \frac{p-1}{2} \) is **even** if \( p \equiv 1 \pmod{4} \) - \( \frac{p-1}{2} \) is **odd** if \( p \equiv 3 \pmod{4} \) Thus, \(-1\) is a quadratic residue mod \( p \) **if and only if** \( p \equiv 1 \pmod{4} \). **Explanation:** The parity of \( \frac{p-1}{2} \) determines whether \((-1)^{\frac{p-1}{2}} \equiv 1 \pmod{p}\), which by Euler's Criterion means \(-1\) is a quadratic residue if and only if \( p \equiv 1 \pmod{4} \). --- ### **Final Answer** **\(-1\) is a quadratic residue modulo \( p \) if and only if \( p \equiv 1 \pmod{4} \).**