Chapter 2: Problem 21
(a) Prove that if \(d \mid n\), then \(2^{d}-1 \mid 2^{n}-1\). [Hint: Use the identity $$ \left.x^{k}-1=(x-1)\left(x^{t-1}+x^{t-2}+\cdots+x+1\right) .\right] $$ (b) Verify that \(2^{35}-1\) is divisible by 31 and 127 .
Short Answer
Expert verified
(a) Proven using identity for overall divisibility. (b) Verified that 31 and 127 divide \(2^{35}-1\).
Step by step solution
01
Understand the Relation
Given that \(d \mid n\), it means \(n = kd\) for some integer \(k\). Thus, we need to show that \(2^d - 1 \mid 2^{kd} - 1\).
02
Use the Hint
Consider the identity: \[x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + \cdots + x + 1).\] Let \(x = 2^d\), then the expression becomes \((2^d)^k - 1 = (2^d - 1)((2^d)^{k-1} + (2^d)^{k-2} + \cdots + 2^d + 1)\). This shows \(2^d - 1\) is a factor of \((2^d)^k - 1 = 2^{kd} - 1\).
03
Conclusion of Part (a)
Since \(2^d - 1\) divides \(2^{kd} - 1\), we've proved that if \(d \mid n\), then \(2^d - 1 \mid 2^n - 1\).
04
Calculate \(2^{35} - 1\)
Calculate \(2^{35} - 1\) to verify its divisibility. Calculate \(2^5 = 32\) and then subsequently calculate powers of 2 up to 35.
05
Check Divisibility by 31
Using Fermat's Little Theorem, which states \(a^{p-1} \equiv 1 \pmod{p}\) for a prime \(p\), check \(2^{30} \equiv 1 \pmod{31}\). Thus \(2^{35} \equiv 2^5 \equiv 32 \equiv 1 \pmod{31}\), confirming \(2^{35} - 1\) is divisible by 31.
06
Check Divisibility by 127
Similarly, since 127 is prime, \(2^{126} \equiv 1 \pmod{127}\). Hence \(2^{35} \equiv 1 \pmod{127}\) must also hold, implying \(2^{35} - 1\) is divisible by 127.
07
Verify the Conclusion
Both 31 and 127 divide \(2^{35} - 1\), which confirms the divisibility as required in part (b).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fermat's Little Theorem
Fermat's Little Theorem is a fundamental principle in number theory that gives us a way to simplify complex calculations involving modular arithmetic. It states that if \( p \) is a prime number and \( a \) is an integer such that \( a ot\equiv 0 \pmod{p} \), then: \[ a^{p-1} \equiv 1 \pmod{p} \] This theorem is particularly useful when checking the divisibility of power expressions by a prime number. In our original exercise, we used Fermat's Little Theorem to verify the divisibility of the expression \( 2^{35} - 1 \) by the primes 31 and 127.
- Firstly, for 31: Since \( 31 \) is prime, applying Fermat's theorem gives us \( 2^{30} \equiv 1 \pmod{31} \). Therefore, \( 2^{35} = 2^{30} \times 2^5 \equiv 1 \times 32 \equiv 1 \pmod{31} \), confirming \( 2^{35} - 1 \) is divisible by 31.
- Secondly, for 127: Similarly, 127 being prime ensures \( 2^{126} \equiv 1 \pmod{127} \), which means \( 2^{35} \equiv 1 \pmod{127} \), thus showing \( 2^{35} - 1 \) is divisible by 127 as well.
Factors of Integers
An integer is divisible by another if there is no remainder upon division. This concept plays a crucial role in understanding factorization and divisibility in mathematics. The divisibility of power expressions, such as \( 2^n - 1 \), can be investigated using identities and divisibility rules. In our exercise, we started with an identity to show that if \( d \mid n \), then \( 2^{d}-1 \mid 2^{n}-1 \). We utilized the hint provided, which is based on factoring a difference of powers:
- The identity given was \( x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + \cdots + x + 1) \).
- We substitute \( x = 2^d \) and \( k \) as \( n/d \), to express \((2^d)^k - 1\).
- It reveals \( (2^d)^k - 1 = (2^d - 1)Q \), where \( Q \) is a polynomial in \( 2^d \), implying that \( 2^d - 1 \) is a factor of \( 2^{n} - 1 \).
Power of Two Expressions
Power of two expressions, like \( 2^n \), exhibit special properties that make them intriguing in mathematics. Understanding these properties requires examining how expressions grow and interact with basic arithmetic.
- Exponential Growth: Powers of two increase exponentially, meaning each subsequent power is double the previous one. This rapid growth can lead to challenging calculations without formulas or shortcuts.
- Divisibility Patterns: As highlighted in the original exercise, the expressions \( 2^n - 1 \) can be factored and analyzed for divisibility by specific integers. Techniques involving Fermat's theorem and polynomial identities simplify these tasks.