Abstract Algebra Theory and Applications - Computer Science ...

66 CHAPTER 3 CYCLIC GROUPSk multiplications by computinga 20 (mod n)a 21 (mod n)a 2k.(mod n).Each step involves squaring the answer obtained in the previous step, dividingby n, and taking the remainder.Example 12. We will compute 271 321 (mod 481). Notice that321 = 2 0 + 2 6 + 2 8 ;hence, computing 271 321 (mod 481) is the same as computing271 20 +2 6 +2 8 ≡ 271 20 · 271 26 · 271 28 (mod 481).So it will suffice to compute 271 2i (mod 481) where i = 0, 6, 8. It is veryeasy to see that271 21 ≡ 73, 441 (mod 481)≡ 329 (mod 481).We can square this result to obtain a value for 271 22 (mod 481):271 22 ≡ (271 21 ) 2 (mod 481)≡ (329) 2 (mod 481)≡ 1, 082, 411 (mod 481)≡ 16 (mod 481).We are using the fact that (a 2n ) 2 ≡ a 2·2n ≡ a 2n+1 (mod n). Continuing, wecan calculate271 26 ≡ 419 (mod 481)andTherefore,271 28 ≡ 16 (mod 481).271 321 ≡ 271 20 +2 6 +2 8 (mod 481)≡ 271 20 · 271 26 · 271 28 (mod 481)≡ 271 · 419 · 16 (mod 481)≡ 1, 816, 784 (mod 481)≡ 47 (mod 481).