Help me fast If m = 2* is a power of 2, explain how you could use repeated squaringto compute am (mod n) for all n. Then apply your method to compute 102(mod 41).If m is not a power of 2, explain how you could use the results of the exerciseabove to compute a™ (mod n) for any n. Then apply your method to compute1726 (mod 44).

Answered: If m = 2* is a power of 2, explain how…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.5: The Binomial Theorem

Problem 49E

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If m = 2* is a power of 2, explain how you could use repeated squaring
to compute am (mod n) for all n. Then apply your method to compute 102
(mod 41).
If m is not a power of 2, explain how you could use the results of the exercise
above to compute a™ (mod n) for any n. Then apply your method to compute
1726 (mod 44).

Expert Solution

Step 1

a) We have m = $2^{k}$ . First we illustrate the method of repeated squaring to compute $a^{m} (m o d n)$ . Then we will compute $10^{32} (m o d 41)$ .
The following steps compute the value of $a^{m} (m o d n) :$

1. Write m as a sum of powers of 2,
$m = u_{0} + u_{1} . 2 + u_{2} . 4 + u_{3} . 8 + . . . + u_{r} . 2^{r}$ ,
where each $u_{i}$ is either 0 or 1. (This is called the binary expansion of m.)

2.Make a table of powers of a modulo m using successive squaring.
$a^{1} \equiv A_{0} (m o d n) a^{2} \equiv {(a^{1})}^{2} = A_{0}^{2} \equiv A_{1} (m o d n) a^{4} \equiv {(a^{2})}^{2} = A_{1}^{2} \equiv A_{2} (m o d n) a^{8} \equiv {(a^{4})}^{2} = A_{2}^{2} \equiv A_{3} (m o d n) . . . a^{2^{r}} \equiv {(a^{2^{r - 1}})}^{2} = A_{r - 1}^{2} \equiv A_{r} (m o d n)$
Note that to compute each line of the table we only need to take the number at the end of
the previous line, square it, and then reduce it modulo m. Also note that the table has r + 1
lines, where r is the highest exponent of 2 appearing in the binary expansion of k in Step 1.

Step 2

3. The product $A_{0}^{u_{0}}, A_{1}^{u_{1}}, A_{2}^{u_{2}}, . . ., A_{r}^{u_{r}} m o d n$ , will be congruent to $a^{m} (m o d n)$ . Note that all of the $u_{i}$ 's are either 0 or 1, so this number is really just the product of those $A_{i}' s$ for which u1 is 1.
We compute
$a^{m} \equiv a^{u_{0} + u_{1} . 2 + u_{2} . 4 + u_{3} . 8 + . . . + u_{r} . 2^{r}} \equiv a^{u_{0}} + {(a^{2})}^{u_{1}} + {(a^{4})}^{u_{2}} + {(a^{8})}^{u_{3}} + . . . + {(a^{2^{r}})}^{u_{r}} \equiv A_{0}^{u_{0}} + A_{1}^{u_{1}} + A_{2}^{u_{2}} + . . . + A_{r}^{u_{r}}$

Now, we compute the value of $10^{32} (m o d 41)$ .
We can 32 as $32 = 0 + 0.2 + 0 . 2^{2} + 0 . 2^{3} + 0 . 2^{4} + 1 . 2^{5}$

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