Chapter 23: Problem 3
Let \(p\) be a prime \(>2\). If \(p \equiv 3(\bmod 4)\), then \((p-1) / 2\) is odd. (b) Let \(p>2\) be a prime such that \(p \equiv 3\) (mod 4). Then there is no solution to the congruence \(x^{2}+1 \equiv 0(\bmod p) .\) [HiNT: Raise both sides of \(x^{2} \equiv-1(\bmod p)\) to the power \((p-1) / 2\), and use Fermat's little theorem.]
Short Answer
Step by step solution
Understanding the Problem
Analyze the Congruence Condition
Set up the Equation for Part (b)
Use Fermat's Little Theorem
Simplify the Powers
Conclusion: No Solutions Exist
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