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1. Solve the congruence x² + 1 = 0 mod 13³.

1. Solve the congruence x² + 1 = 0 mod 13³.

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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1. Solve the congruence 2. Solve the congruence x² + 1 = 0 mod 13³. x³ - 2x² - 7x +3=0_mod 7². 3. Suppose that n is an odd positive integer and write o(n) a b n with ged(a, b) = 1. Show that the largest prime factor of b is the largest prime factor of n 4. Find the smallest positive odd integer n such that o(n) 7680 n 12121 You should do this by a mathematical analysis not a brute force computer search. Hint: Use Problem 3. e.g., R₂ = 11, R3 = 111, R4 = 5. Suppose that a is a positive integer relatively prime to 10. Show that a divides infinitely many 'repunits', i.e., numbers of the form Rn = 1111, etc. Hint: Rn 11…….11. 10 - 1 10 1 Hint: The case where 3 does not divide a is easier. = n
Transcribed Image Text:1. Solve the congruence 2. Solve the congruence x² + 1 = 0 mod 13³. x³ - 2x² - 7x +3=0_mod 7². 3. Suppose that n is an odd positive integer and write o(n) a b n with ged(a, b) = 1. Show that the largest prime factor of b is the largest prime factor of n 4. Find the smallest positive odd integer n such that o(n) 7680 n 12121 You should do this by a mathematical analysis not a brute force computer search. Hint: Use Problem 3. e.g., R₂ = 11, R3 = 111, R4 = 5. Suppose that a is a positive integer relatively prime to 10. Show that a divides infinitely many 'repunits', i.e., numbers of the form Rn = 1111, etc. Hint: Rn 11…….11. 10 - 1 10 1 Hint: The case where 3 does not divide a is easier. = n
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