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3. Solve the system of congruences Show the steps used. 3x = 2 7x3 8x 7 mod 7 mod 11 mod 19.

3. Solve the system of congruences Show the steps used. 3x = 2 7x3 8x 7 mod 7 mod 11 mod 19.

Advanced Engineering Mathematics
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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[Number Theory] How do you solve question 3?

1. Find all solutions of the congruence 5x² + 3x² + x + 3 = 0 mod 30. You should do this by factoring the modulus and using the CRT, as explained in class. 2. Recall that a number m is said to be square free if d² | m for d≥ 1 implies that d = 1. Equivalently, m is not divisible by the square of any prime p. Show that there are infinitely many integers n such that each of the numbers n, n+1, n+2 and n+ 3 not square free. 3. Solve the system of congruences 3x2 mod 7 7x3 mod 11 8x 7 mod 19. Show the steps used. 4. (i) For an odd number n, suppose that 2" # 2 mod n. Can n be a prime? Explain. This is called the base 2 primality test. (ii) Using the fact that 187 = 11-17, show that 2186174 mod 187. This shows that 187 fails the base 2 primality test. (iii) Compute 2561 mod 561. Use Fermat's little theorem and hand calculations. Does 561 pass or fail the base 2 primality test? You may use the prime factorization of 561.
Transcribed Image Text:1. Find all solutions of the congruence 5x² + 3x² + x + 3 = 0 mod 30. You should do this by factoring the modulus and using the CRT, as explained in class. 2. Recall that a number m is said to be square free if d² | m for d≥ 1 implies that d = 1. Equivalently, m is not divisible by the square of any prime p. Show that there are infinitely many integers n such that each of the numbers n, n+1, n+2 and n+ 3 not square free. 3. Solve the system of congruences 3x2 mod 7 7x3 mod 11 8x 7 mod 19. Show the steps used. 4. (i) For an odd number n, suppose that 2" # 2 mod n. Can n be a prime? Explain. This is called the base 2 primality test. (ii) Using the fact that 187 = 11-17, show that 2186174 mod 187. This shows that 187 fails the base 2 primality test. (iii) Compute 2561 mod 561. Use Fermat's little theorem and hand calculations. Does 561 pass or fail the base 2 primality test? You may use the prime factorization of 561.
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