Can you do #4? Example 4.4.5. If p =has a solution.The proof is an application of the theorems of Fermat and Lagrange.Write p = 4k +1. By Fermat's Theorem, the equation x4k = 1 (mod p) hasAk solutions in a complete residue system. We can factora4k – 1 = (22k – 1)(x2k + 1) (mod p).- 1 = 0 (mod p) or aEvery root of the left-hand side is either a root of x2kroot of r2k.(mod p) has exactly 2k solutions. Therefore, the congruence x2k + 1 = 0(mod p) has 2k solutions. If a is a root, then (ak)2 = -1 (mod p), so ak isa solution to the congruence x² = -1 (mod p).The assertion of this example has many consequences. It can be used toshow that a prime p of the form 4k + 1 can be written as a sum of 2 squares.(See Section 14.4 for a proof.)+1 = 0 (mod p). By Corollary 4.4.3, the congruence x2k –1 = 0Exercises for Section 4.41. Find the four solutions of x = 1 (mod 13).2. Suppose p is prime and d p - 1. Show that xd - 1 = 0 (mod p) hasexactly (d, p – 1) solutions. Conclude that if 4 | p- 1, then x2 = -1(mod p) has two solutions, and if 4 p- 1, then x2 = -1 (mod p) has nosolutions.3. Suppose p =xk = -1 (mod p) have?6k + 1 is prime. How many solutions does the congruence4. Let p be prime.(a) If a 1(b) If a z -1 (mod p), show that 1 – a + a? - a³ + · . . + aP-1 = 1(mod p).mod p), show that 1 + a + ...+ aP-1 = 1 (mod p).5. Let p be an odd prime and letf (x) = (x – 1)(x – 2) . . - (x – (p – 1))= rP-1+ ap-2xP-2+...+ a1x + ao.

Name: Can you do #4? Example 4.4.5. If p =has a solution.The proof is an app
Uploaded: 2024-03-20T07:06:41.000Z
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Description: Can you do #4? Example 4.4.5. If p =has a solution.The proof is an application of the theorems of Fermat and Lagrange.Write p = 4k +1. By Fermat's Theorem, the equation x4k = 1 (mod p) hasAk solutions in a complete residue system. We can factora4k –