Use the method described in Sect. 2.8.1 to find square roots modulo the following composite moduli.(d) Find eight square roots of 813 modulo 868.(I've included the examples gone over in Sect 2.8.1) Example 2.27. A square root of \( a = 2201 \) modulo the prime \( p = 4127 \) is \[ b \equiv a^{(p+1)/4} \equiv 2201^{4128/4} \equiv 2201^{1032} \equiv 3718 \pmod{4127}. \]To see that \( a \) does indeed have a square root modulo 4127, we simply square \( b \) and check that \( 3718^2 \equiv 13823524 \equiv 2201 \pmod{4127}. \)Suppose now that we want to compute a square root modulo \( m \), where \( m \) is not necessarily a prime. An efficient method is to factor \( m \), compute the square root modulo each of the prime (or prime power) factors, and then combine the solutions using the Chinese remainder theorem. An example makes the idea clear. Example 2.28. We look for a solution to the congruence \[ x^2 \equiv 197 \pmod{437}. \tag{2.11} \]The modulus factors as \( 437 = 19 \cdot 23 \), so we first solve the two congruences \[ y^2 \equiv 197 \pmod{19} \quad \text{and} \quad z^2 \equiv 197 \pmod{23}. \]Since both 19 and 23 are congruent to 3 modulo 4, we can find these square roots using Proposition 2.26 (or by trial and error). In any case, we have \[ y \equiv \pm 8 \pmod{19} \quad \text{and} \quad z \equiv \pm 6 \pmod{23}. \]We can pick either 8 or \(-8\) for \( y \) and either 6 or \(-6\) for \( z \). Choosing the two positive solutions, we next use the Chinese remainder theorem to solve the simultaneous congruences \[ x \equiv 8 \pmod{19} \quad \text{and} \quad x \equiv 6 \pmod{23}. \tag{2.12} \]We find that ## 2.8.1 Solving Congruences with Composite ModuliIt is usually easiest to solve a congruence with a composite modulus by first solving several congruences modulo primes (or prime powers) and then fitting together the solutions using the Chinese remainder theorem. We illustrate the principle in this section by discussing the problem of finding square roots modulo *m*. It turns out that it is relatively easy to compute square roots modulo a prime. Indeed, for primes congruent to 3 modulo 4, it is extremely easy to find square roots, as shown by the following proposition.**Proposition 2.26.** Let *p* be a prime satisfying *p* ≡ 3 (mod 4). Let *a* be an integer such that the congruence *x²* ≡ *a* (mod *p*) has a solution, i.e., such that *a* has a square root modulo *p*. Then## 2.8. The Chinese Remainder Theorem**b** ≡ *a*^((p+1)/4) (mod *p*)is a solution, i.e., it satisfies *b²* ≡ *a* (mod *p*). (*N.B.* This formula is valid only if *a* has a square root modulo *p*. In Sect. 3.9 we will describe an efficient method for checking which numbers have square roots modulo *p*.)**Proof.** Let *g* be a primitive root modulo *p*. Then *a* is equal to some power of *g*, and the fact that *a* has a square root modulo *p* means that *a* is an even power of *g*, say *a* ≡ *g²k* (mod *p*). (See Exercise 2.5.) Now we compute*b²* ≡ *a*^((p+1)/2) (mod *p*) definition of *b*, ≡ (*g²k*)^((p+1)/2) (mod *p*) since *a* ≡ *g²k* (mod *p*), ≡ *g*^(p+1)k (mod *p*) ≡ *g²k*+ (p−1)k

Using Proposition 2.26 we solve this problem

Answered: Use the method described in Sect. 2.8.1…

Use the method described in Sect. 2.8.1 to find square roots modulo the following composite moduli. (d) Find eight square roots of 813 modulo 868. (I've included the examples gone over in Sect 2.8.1)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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