Section 41 Suppose that a and b are integers with a 3 mod 11

(Section 4.1) Suppose that a and b are integers with a 3 (mod 11) and b 6 (mod 11). Find an integer c

(with0 c < 11) for which b2 5a + c (mod11).

(Section 4.1) Show that if n is an odd integer, then n2 1 (mod 8).

Solution

b^2=6^2=36=3 mod 11

5a=15=4 mod 11

b^2-5a=3-4=-1=10 mod 11

Hence, c=10 is one such integer

n is odd

Case 1:n=4k+1

n^2=16k^2+8k+1=1 mod 8

Case 2:n=4k-1

n^2=16k^2-8k+1=1 mod 8

Hence proved