Introduction to Abstract Algebra, Solutions Manual, 4th Edition

1.1 Induction

1. In each case we give the equation that makes p_k imply p_k+1.

a. k(2k−1)+(4k+1)=2k²+3k+1=(k+1)(2k+1)

c.

e.

g. 7

i.

2. In each case we give the inequality that makes p_k imply p_k+1.

a. 2^k+1=2·2^k>2·k≥k+1.

c. If , then provided k+1≤2^2k+1. This latter inequality follows, again by induction on k≥1, because 2^2k+3=4·2^2k+1≥4(k+1)≥k+2.

e. .

3. In each case we give the calculation that makes p_k imply p_k+1.

a. If k³+(k+1)³+(k+2)³=9m, then (k+1)³ + (k+2)³ + (k+3)³ = 9m−k³ + (k+3)³ = 9m+9k² + 27k + 27.

c. If 3^2k+1+2^k+2=7m, then

5. If 3^3k+1=7m where k is odd, then passing to ...