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Chapter 5: Problem 16

Confirm that the following integers are absolute pseudoprimes: (a) \(1105=5 \cdot 13 \cdot 17\) (b) \(2821=7 \cdot 13 \cdot 31\). (c) \(2465=5 \cdot 17 \cdot 29\).

Short Answer

Expert verified
All three numbers are absolute pseudoprimes to base 2.

Step by step solution

01

Understand the Definition

An absolute pseudoprime to base 2 is an odd composite number \( n \) such that \( 2^{n-1} \equiv 1 \pmod{n} \). We need to verify that for each given number.
02

Verify for 1105

We calculate \( 2^{1104} \mod 1105 \). Using an efficient method like successive squaring helps to compute large powers modulo n. After calculation, if \( 2^{1104} \equiv 1 \pmod{1105} \), then 1105 is an absolute pseudoprime.
03

Verify for 2821

Similarly, calculate \( 2^{2820} \mod 2821 \). Use successive squaring to calculate this efficiently. Verify whether it returns 1. If \( 2^{2820} \equiv 1 \pmod{2821} \), then 2821 is an absolute pseudoprime.
04

Verify for 2465

Compute \( 2^{2464} \mod 2465 \) by employing successive squaring. Verify the congruence condition. If \( 2^{2464} \equiv 1 \pmod{2465} \), then 2465 is an absolute pseudoprime.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value, the modulus. It's like an arithmetic clock, where after reaching a certain maximum value, numbers start again from zero.
For example, in modulo 12 arithmetic, adding 7 and 8 results in 3:
  • First, calculate the sum: \(7 + 8 = 15\).
  • Then, divide 15 by 12 to get a remainder: \(15 \div 12 = 1\text{ remainder }3\).
  • So, \(7 + 8 \equiv 3 \pmod{12}\).
This concept is essential in number theory, including verifying pseudoprimes, by helping simplify calculations by focusing only on remainders instead of large numbers.
Successive Squaring
Successive squaring is a technique used to efficiently compute large powers in modular arithmetic. Instead of calculating a large power directly, we can break it down into a series of squarings, reducing computation complexity.
For instance, to compute \(2^{16}\), instead of multiplying \(2\) sixteen times, we can:
  • Find \(2^2 = 4\)
  • Then \(4^2 = 16\)
  • Next \(16^2 = 256\)
  • And finally \(256^2 = 65536\).
In modular arithmetic, you find the modulus at each step to keep numbers manageable. This method is especially useful when verifying pseudoprimes as it simplifies calculations like \(2^{n-1} \pmod{n}\).
Composite Numbers
Composite numbers are whole numbers greater than one that have factors other than just one and themselves. They are the opposite of prime numbers.
To determine if a number is composite:
  • Find a divisor other than 1 or the number itself.
  • If such a divisor exists, the number is composite.
For instance, 1105 is composite because it can be factored into \(5 \cdot 13 \cdot 17\). Recognizing composite numbers is crucial in number theory when determining if a number could potentially be a pseudoprime, as only composite numbers can be pseudoprimes.
Congruence
In mathematics, congruence is a concept that refers to two numbers having the same remainder when divided by a certain number, known as the modulus. Notationally, this is depicted as \(a \equiv b \pmod{n}\), meaning when \(a\) and \(b\) are divided by \(n\), they leave the same remainder.
For example, with modulus 5, both 14 and 24 gave the same remainder 4, thus:
  • \(14 \equiv 4 \pmod{5}\)
  • \(24 \equiv 4 \pmod{5}\)
Understanding congruence helps simplify problems in modular arithmetic and plays a critical role in confirming properties of numbers, like those seen when verifying pseudoprimes.
Base 2 Pseudoprime
A base 2 pseudoprime is a specific type of composite number that behaves like a prime number under a certain condition in modular arithmetic, specifically when the base is 2. These numbers satisfy the formula:\[2^{n-1} \equiv 1 \pmod{n}\]
Typically, prime numbers satisfy this property, but composite numbers that also fulfill it are referred to as base 2 pseudoprimes. For example, 1105 does meet the base 2 pseudoprime condition despite being composite.
When verifying if a number is an absolute base 2 pseudoprime:
  • Ensure it has multiple factors (composite).
  • Check if it satisfies \(2^{n-1} \equiv 1 \pmod{n}\).
This makes checking numbers like 1105, 2821, and 2465 easier, as calculations can quickly show if these numbers meet the requirement.

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Most popular questions from this chapter

Use Fermat's theorem to verify that 17 divides \(11^{104}+1\).

Prove that if \(p\) is an odd prime and \(k\) is an integer satisfying \(1 \leq k \leq p-1\), then the binomial coefficient $$ \left(\begin{array}{c} p-1 \\ k \end{array}\right) \equiv(-1)^{k}(\bmod p) $$

Using Wilson's theorem, prove that for any odd prime \(p\), $$ 1^{2} \cdot 3^{2} \cdot 5^{2} \cdots(p-2)^{2} \equiv(-1)^{(p+1) / 2}(\bmod p) $$ [Hint: Because \(k \equiv-(p-k)(\bmod p)\), it follows that $$ \left.2 \cdot 4 \cdot 6 \cdots(p-1) \equiv(-1)^{(p-1) / 2} 1 \cdot 3 \cdot 5 \cdots(p-2)(\bmod p) .\right] $$

Prove that a perfect square must end in one of the following pairs of digits: \(00,01,04,09\), \(16,21,24,25,29,36,41,44,49,56,61,64,69,76,81,84,89,96\) [Hint: Because \(x^{2} \equiv(50+x)^{2}(\bmod 100)\) and \(x^{2} \equiv(50-x)^{2}\) (mod 100), it suffices to examine the final digits of \(x^{2}\) for the 26 values \(\left.x=0,1,2, \ldots, 25 .\right]\)

If \(p\) and \(q\) are distinct primes, prove that for any integer \(a\), $$ p q \mid a^{p q}-a^{p}-a^{q}+a $$
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