A Book of Abstract Algebra

Charles C. Pinter

Chapter 23 ELEMENTS OF NUMBER THEORY - all with Video Answers

Educators

Section 1

Problem 1

Solving Single Congruences
For each of the following congruences, find $m$ such that the congruence has a unique solution modulo $m$. If there is no solution, write "none."
(a) $60 x \equiv 12(\bmod 24)$
(b) $42 x \equiv 24(\bmod 30)$
(c) $49 x \equiv 30(\bmod 25)$
(d) $39 x \equiv 14(\bmod 52)$
(e) $147 x=47(\bmod 98)$
(f) $39 x \equiv 26(\bmod 52)$

Nick Johnson

Numerade Educator

03:06

Problem 2

Solve the following linear congruences:
(a) $12 x \equiv 7(\bmod 25)$
(b) $35 x \equiv 8(\bmod 12)$
(c) $15 x \equiv 9(\bmod 6)$
(d) $42 x \equiv 12(\bmod 30)$
(e) $147 x \equiv 49(\bmod 98)$
(f) $39 x \equiv 26(\bmod 52)$

James Chok

Numerade Educator

01:28

Problem 3

Explain why $2 x^{2} \equiv 8(\bmod 10)$ has the same solutions as $x^{2}=4(\bmod 5)$.
(b) Explain why $x \equiv 2(\bmod 5)$ and $x \equiv 3(\bmod 5)$ are all the solutions of $2 x^{2} \equiv 8$
$($ mod 10$)$.

Heather Zimmers

Numerade Educator

04:07

Problem 4

Solve the following quadratic congruences (if there is no solution, write "none"):
(a) $6 x^{2} \equiv 9(\bmod 15)$
(b) $60 x^{2} \equiv 18(\bmod 24)$
(c) $30 x^{2} \equiv 18(\bmod 24)$
(d) $4(x+1)^{2} \equiv 14(\bmod 10)$
(f) $3 x^{2}-6 x+6 \equiv 0(\bmod 15)$

Erika Bustos

Numerade Educator

03:35

Problem 5

Solve the following congruences:
(a) $x^{4} \equiv 4(\bmod 6)$
$\left(\right.$ b) $2(x-1)^{4} \equiv 0(\bmod 8)$
(c) $x^{3}+3 x^{2}+3 x+1 \equiv 0(\bmod 8)$
$\left(\right.$ d) $x^{4}+2 x^{2}+1 \equiv 4(\bmod 5)$

M Hassan Anwar

Numerade Educator

03:38

Problem 6

Solve the following Diophantine equations (if there is no solution, write "none"):
(a) $14 x+15 y=11$
(b) $4 x+5 y=1$
(c) $21 x+10 y=9$
(d) $30 x^{2}+24 y=18$

Erika Bustos

Numerade Educator