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2. Use the method of squaring to compute 51711 mod 1911.

2. Use the method of squaring to compute 51711 mod 1911.

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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[Pseudoprimes] How do you solve #2? The second picture is for context

1. Show that 25 is a strong pseudoprime base 7, i.e., passes Miller's test base 7. 2. Use the method of squaring to compute 51711 mod 1911. 3. (i) For a prime p suppose that n = 2º - 1 is not a prime. Show that n is a pseudoprime base 2. (ii) Show every composite Fermat number Fm 4. Show that if = n 22m + 1 is a pseudoprime base 2. a²p 1 a² - 1 where a > 1 is an integer and p is an odd prime with pła(a²-1), then n is a pseudoprime base a. Hint: First show that 2p | (n − 1). Then consider an−¹ – 1. Note that this shows that there are infinitely many pseudoprimes base a for any a.
Transcribed Image Text:1. Show that 25 is a strong pseudoprime base 7, i.e., passes Miller's test base 7. 2. Use the method of squaring to compute 51711 mod 1911. 3. (i) For a prime p suppose that n = 2º - 1 is not a prime. Show that n is a pseudoprime base 2. (ii) Show every composite Fermat number Fm 4. Show that if = n 22m + 1 is a pseudoprime base 2. a²p 1 a² - 1 where a > 1 is an integer and p is an odd prime with pła(a²-1), then n is a pseudoprime base a. Hint: First show that 2p | (n − 1). Then consider an−¹ – 1. Note that this shows that there are infinitely many pseudoprimes base a for any a.
Example 4.5 Let us apply the base 2 test to the integer n = 341. Computing 2341 mod (341) is greatly simplified by noting that 2¹0 1024 1 mod (341), so 2341 = (2¹0) 342 = 2 mod (341), and 341 has passed the test. However 341 = 11.31, so it is not a prime but a pseudo-prime. (In fact, knowing this factorisation in advance, one could 'cheat' in the base 2 test to avoid large computations: since 11 and 31 are primes, Theorem 4.3 gives 2¹⁰ = 1 mod (11) and 2³⁰ = 1 mod (31), which easily imply that 2341 – 2 is divisible by both 11 and 31, and hence by 341.) By checking that all composite numbers n < 341 fail the base 2 test, one can show that 341 is the smallest pseudo-prime. =
Transcribed Image Text:Example 4.5 Let us apply the base 2 test to the integer n = 341. Computing 2341 mod (341) is greatly simplified by noting that 2¹0 1024 1 mod (341), so 2341 = (2¹0) 342 = 2 mod (341), and 341 has passed the test. However 341 = 11.31, so it is not a prime but a pseudo-prime. (In fact, knowing this factorisation in advance, one could 'cheat' in the base 2 test to avoid large computations: since 11 and 31 are primes, Theorem 4.3 gives 2¹⁰ = 1 mod (11) and 2³⁰ = 1 mod (31), which easily imply that 2341 – 2 is divisible by both 11 and 31, and hence by 341.) By checking that all composite numbers n < 341 fail the base 2 test, one can show that 341 is the smallest pseudo-prime. =
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