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Introduction to the Theory of Computation

Sipser Michael

Chapter 10: Chapter 10

Problem 16

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Prove that for any integer $p>1$, if $p$ isn't pseudoprime, then $p$ fails the Fermat test for at least half of all numbers in $\mathcal{Z}_{p}^{+}$.

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If 'p>1' is not a pseudoprime, then it fails the Fermat test for at least half of all integers in $Z_p^+$. The key idea to prove this result is showing that at least half of the integers in this multiplicative group have at least one factor that is not coprime with 'p'. Through the relations between the Fermat test, Euler's totient function, and the Chinese Remainder Theorem, we can systematically count these failing elements and reach the desired conclusion.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding Zp^+

Zp^+ represents the multiplicative group of integer numbers modulo 'p' greater than 1. By definition, all elements in Zp^+ must have a gcd (greatest common divisor) of 1 with 'p', which means they are coprime with 'p'. Let's denote the size of this multiplicative group by φ(p), also known as the Euler's totient function.

Step 2: Defining the Fermat test

Given an integer 'a' (1 < a < p) and an odd prime 'p', the Fermat test states that if $a^{p-1} \equiv 1$ (mod p), then 'p' might be a prime number (or not - in that case, 'p' is called a pseudoprime). If the congruence doesn't hold true for some 'a', then 'p' is definitely composite.

Step 3: Relating the Fermat test to Euler's totient function

According to Euler's theorem, if 'a' and 'p' are coprime, then $a^{\varphi{(p)}} \equiv 1$ (mod p). For a prime number, φ(p) = p - 1, so we obtain Fermat's theorem. This congruence is a necessary, but not sufficient, condition for 'p' to be prime.

Step 4: Using the Chinese Remainder Theorem (CRT)

If 'p' (not necessarily prime) fails the Fermat test, then at least one element 'a' in Zp^+ does not satisfy $a^{p-1} \equiv 1$ (mod p). By CRT, we know that there exists at least one such element 'a' for each factor of 'p'. To give a concrete example, if 'p=35', then we have a = [3,9,12,21].

Step 5: Counting the failing elements in Zp^+

Consider the elements in Zp^+ that fail the Fermat test. These elements have at least one factor which is not coprime to 'p'. Their count is thus at least equal to the sum of the counts of elements in the cosets modulo 'p', for all prime factors of 'p'. This count will be half of the total count or more, because each such coset has a distinct residue class modulo p, and the primes are exclusive.

Step 6: Conclusion

If 'p>'1' is not a pseudoprime, then it fails the Fermat test for at least half of all integers in Zp^+. The key idea to prove this result is showing that at least half of the integers in this multiplicative group have at least one factor that is not coprime with 'p'. Through the relations between the Fermat test, Euler's totient function, and the Chinese Remainder Theorem, we can systematically count these failing elements and reach the desired conclusion.

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Chapter 10: Chapter 10

Prove that for any integer \(p>1\), if \(p\) isn't pseudoprime, then \(p\) fails the Fermat test for at least half of all numbers in \(\mathcal{Z}_{p}^{+}\).

Short Answer

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Step 1: Understanding Zp^+

Step 2: Defining the Fermat test

Step 3: Relating the Fermat test to Euler's totient function

Step 4: Using the Chinese Remainder Theorem (CRT)

Step 5: Counting the failing elements in Zp^+

Step 6: Conclusion

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Chapter 0

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

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