Question

Use criteria shown in the proof of Theorem 7.5 .10 to find three primitive roots modulo 25. That is, find three generators of Z25∗. Theore 7.5. 0 Let p be an odd prime, and bt k∈Z>0. Then Zp∗∗ is a cyclic group. Let a genente Zp∗. Then a+p≡a(modp) is also a gerenator. If ap−1≡1(modp2) then (a+p)p−1≡ap−1+p(p−1)ap−2≡1+p(p−1)ap−2(modp2). Since p∤a, this shows (a+p)p−1≡1(modp2). So, we have ap−1≡1(modp) and ap−1=1(modp2). with some heary cemputations, we can show o(a)=φ(pk)=∣∣Zp∗∗∣∣.

Answer 1

To find three primitive roots modulo 25, we can use the criteria shown in the proof of Theorem 7.5.10. We find that 2, 7, and 12 are generators of Z25∗.

Step-by-step explanation:

In order to find three primitive roots modulo 25, we can use the criteria shown in the proof of Theorem 7.5.10. Theorem 7.5.10 states that if p is an odd prime and a is a generator of Zp∗, then a+p≡a(modp) is also a generator. Additionally, if ap−1≡1(modp2) but ap−1=1(modp), then (a+p)p−1≡1(modp2).

So, let's find three generators of Z25∗. First, we'll find a generator of Z5∗. Since 5 is a prime number, all non-zero residues from 1 to 4 inclusive are generators of Z5∗. Let's choose 2 as our generator.

Next, we'll find a generator of Z25∗. We know that 2 is a generator of Z5∗. Using the criteria from Theorem 7.5.10, we can find a generator of Z25∗ by finding a number that satisfies the conditions ap−1≡1(modp2) and ap−1≠1(modp). By trying different values, we find that 7 is a generator of Z25∗ because 7^4 ≡ 1 (mod 5^2) but 7^4 ≠ 1 (mod 5).

We now have two generators: 2 and 7. To find the third generator, we can use the fact that if a is a generator, then a+p is also a generator. Adding 5 to 2 and 7, we get 7 and 12 as two more generators of Z25∗.