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If \(n=p q=274279\) and \(\phi(n)=272376\), find the primes \(p\) and \(q\). [Hint: Note that $$ \begin{aligned} &p+q=n-\phi(n)+1 \\ &\left.p-q=\left[(p+q)^{2}-4 n\right]^{1 / 2} .\right] \end{aligned} $$

Short Answer

Expert verified
The primes are \(p = 1747\) and \(q = 157\).

Step by step solution

01

Understanding the problem

We need to find the prime factors \(p\) and \(q\) of \(n=274279\) given that the Euler's Totient function \(\phi(n)=272376\). We have the hint that \(p+q=n-\phi(n)+1\) and \(p-q=\sqrt{(p+q)^2 - 4n}\).
02

Compute \(p+q\)

Using the formula \(p+q=n-\phi(n)+1\), calculate \(p+q = 274279 - 272376 + 1 = 1904\).
03

Compute \((p+q)^2\)

Calculate \((p+q)^2 = 1904^2 = 3625216\).
04

Calculate \(4n\)

Calculate \(4n = 4 \times 274279 = 1097116\).
05

Compute \((p+q)^2 - 4n\)

Subtract \(4n\) from \((p+q)^2\) to get \(3625216 - 1097116 = 2528100\).
06

Compute \(p-q\)

Calculate \(p-q = \sqrt{2528100}\). Since \(2528100\) is a perfect square, \(p-q = 1590\).
07

Solve for \(p\) and \(q\)

Set up the system of equations: \(p+q=1904\) and \(p-q=1590\). Add these equations to solve for \(p\): \(2p=3494\) so \(p=1747\). Subtract the second equation from the first to solve for \(q\): \(2q=314\) so \(q=157\).
08

Verify Solution

Verify by computing \(p \times q = 1747 \times 157 = 274279\) which equals \(n\). Also, verify \(\phi(n) = (p-1)(q-1) = 1746 \times 156 = 272376\), which matches the given value of \(\phi(n)\).

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Key Concepts

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

Euler's Totient Function
Euler's Totient Function, often denoted as \( \phi(n) \), is an important concept in number theory. It helps determine the number of integers up to a given integer \( n \) that are coprime to \( n \).
For a number \( n \) that can be expressed as a product of two prime numbers \( p \) and \( q \), as in the given problem, the totient function formula is \( \phi(n) = (p-1)(q-1) \).
This function is crucial for problems involving factorization of numbers using prime numbers, like the given exercise.
It lays the groundwork for cryptosystems like RSA, where knowing the totient is important for encryption and decryption processes. To utilize \( \phi(n) \) efficiently, it's key to recognize the properties of prime factors:
  • For any prime \( p \), \( \phi(p) = p-1 \).
  • The product formula for \( \phi(n) = \phi(p) \cdot \phi(q) \) holds if \( n = p \cdot q \).
Quadratic Equations
Quadratic equations appear frequently in solving problems related to number theory, like the current one where the equations \( p+q \) and \( (p+q)^2-4n \) are utilized. A quadratic equation is typically expressed in the form \( ax^2 + bx + c = 0 \).
This exercise indirectly requires solving a system akin to quadratic equations.Given:
  • \( p+q \) simplifies to one equation based on the characteristics of \( n \) and \( \phi(n) \).
  • \( p-q = \sqrt{(p+q)^2 - 4n} \), another equation, resembles working out quadratic solutions.
Combining these equations can help find solutions for \( p \) and \( q \). The methodology follows the techniques used in solving quadratic systems with variables transformed into simpler expressions for easier interpretation.
Perfect Squares
Understanding perfect squares helps immensely when solving the quadratic derivation in the exercise. A perfect square is a number that is the square of an integer. In the problem, the difference calculated \( \sqrt{2528100} = 1590 \) is a perfect square, greatly simplifying the problem.Here's why recognizing perfect squares matters:
  • Direct calculation of square roots is simpler when dealing with perfect squares.
  • Perfect squares reveal hidden integer solutions in equations, helping solve for variables like \( p \) and \( q \).
  • Solving \( p-q \) uses the square root of \( 2528100 \), showing it is crucial to understanding its perfect square nature.
Being attentive to perfect squares and their properties allows you to simplify and solve equations that might otherwise seem more complex.

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Most popular questions from this chapter

If the Cacsar cipher produced KDSSB ELUWKGDB, what is the plaintext message?

Find the unique solution of each of the following superincreasing knapsack problems: (a) \(118=4 x_{1}+5 x_{2}+10 x_{3}+20 x_{4}+41 x_{5}+99 x_{6}\). (b) \(51=3 x_{1}+5 x_{2}+9 x_{3}+18 x_{4}+37 x_{5}\). (c) \(54=x_{1}+2 x_{2}+5 x_{3}+9 x_{4}+18 x_{5}+40 x_{6}\).

In a lengthy ciphertext message, sent using a linear cipher \(C \equiv a P+b\) (mod 26), the most frequently occurring letter is \(\mathrm{Q}\) and the second most frequent is J. (a) Break the cipher by determining the values of \(a\) and \(b\). [Hint: The most often used letter in English text is \(\mathrm{E}\), followed by \(\mathrm{T}\).] (b) Write out the plaintext for the intercepted message WCPQ JZQO MX.

(a) A linear cipher is defined by the congruence \(C=a P+b(\bmod 26)\), where \(a\) and \(b\) are integers with \(\operatorname{gcd}(a, 26)=1\). Show that the corresponding decrypting congruence is \(P \equiv a^{\prime}(C-b)(\bmod 26)\), where the integer \(a^{\prime}\) satisfies \(a a^{\prime} \equiv 1(\bmod 26)\). (b) Using the linear cipher \(C=5 P+11\) (mod 26), encrypt the message NUMBER THEORY IS EASY. (c) Decrypt the message RXQTGUHOZTKGHFJKTMMTG, which was produced using the linear cipher \(C=3 P+7(\bmod 26)\).

A plaintext message expressed in Baudot code has been converted by the Verman cipher into the string $$ 110001110000111010100101111111 $$ If it is known that the key used for encipherment was $$ 011101011001011110001001101010 $$ recover the message in its alphabetic form.

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