Let p be prime. (a) If a 1 (mod p), show that 1+ a + ...+ aP-1 = 1 (mod p). (b) If a -1 (mod p), show that 1 – a + a? (mod p). – a³ + ...+ aP¬1 = 1
Let p be prime. (a) If a 1 (mod p), show that 1+ a + ...+ aP-1 = 1 (mod p). (b) If a -1 (mod p), show that 1 – a + a? (mod p). – a³ + ...+ aP¬1 = 1
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 16E
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Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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Can you do #4?

Transcribed Image Text: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) has
Ak solutions in a complete residue system. We can factor
a4k – 1 = (22k – 1)(x2k + 1) (mod p).
- 1 = 0 (mod p) or a
Every root of the left-hand side is either a root of x2k
root 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 is
a solution to the congruence x² = -1 (mod p).
The assertion of this example has many consequences. It can be used to
show 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 = 0
Exercises for Section 4.4
1. 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) has
exactly (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 no
solutions.
3. Suppose p =
xk = -1 (mod p) have?
6k + 1 is prime. How many solutions does the congruence
4. 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 let
f (x) = (x – 1)(x – 2) . . - (x – (p – 1))
= rP-1
+ ap-2xP-2
+...+ a1x + ao.
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