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Chapter 5: Problem 1

Use Fermat's method to factor each of the following numbers: (a) \(2279 .\) (b) 10541 . (c) 340663 [Hint: The smallest square just exceeding 340663 is \(584^{2}\).]

Short Answer

Expert verified
(a) 43 x 53; (b) 83 x 127; (c) Difficult, check closer integer calculations.

Step by step solution

01

Understanding Fermat's Factorization Method

Fermat's method is based on the representation of an odd integer as a difference of two squares. For a number \(N\), find \(x\) such that \(x^2 - N = y^2\). Then \(N = (x - y)(x + y)\). We'll find values of \(x\) and \(y\) for each number given.
02

Step 2(a): Factor 2279 Using Fermat's Method

Start with \(x = \lceil \sqrt{2279} \rceil = 48\). Calculate \(x^2 - 2279\) to see if it's a perfect square:- \(48^2 = 2304\) and \(2304 - 2279 = 25\), which is a perfect square (\(y = 5\)).Therefore, \(2279 = (48 - 5)(48 + 5) = 43 \times 53\).
03

Step 2(b): Factor 10541 Using Fermat's Method

Start with \(x = \lceil \sqrt{10541} \rceil = 103\). Calculate \(x^2 - 10541\) to find if it's a perfect square:- \(103^2 = 10609\) and \(10609 - 10541 = 68\) (not a square).- Increment \(x\) to 104, then \(104^2 = 10816\).- \(10816 - 10541 = 275\) (not a square).- Increment \(x\) to 105, then \(105^2 = 11025\).- \(11025 - 10541 = 484\) (\(22^2\)).Thus, \(10541 = (105 - 22)(105 + 22) = 83 \times 127\).
04

Step 2(c): Factor 340663 Using Fermat's Method

Given hint: Smallest square exceeding 340663 is \(584^2\). Start with \(x = 584\). Calculate \(x^2 - 340663\):- \(584^2 = 341056\) and \(341056 - 340663 = 393\), increment \(x\).- Increment \(x\) to 585, then \(585^2 = 342225\).- \(342225 - 340663 = 1562\) (not a square), continue incrementing.- Continue incrementing through trial and error until we find:- At \(x = 591\), \(591^2 = 349281\) and \(349281 - 340663 = 8618\), check this again as it doesn't work as a perfect square.Continue the incrementing process until discovery:- At \(x = 562\), \(562^2 = 316834\) and \(\sqrt{340663 - 316834} = \sqrt{23829}\) doesn't form a square.However, finding an error in setup seems optimal as similar perfect square closest to smaller values had computation errors. Should be rechecked with a correction in the base trial, indicating \(593^2 - 340663 = 386792, 33104\).Meaning eventually solving correct.However reconfirm in requiring detailed breakdown more depending on early tasks would require possible check with computer or another heightens factor due to complex number.

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Key Concepts

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

Difference of Squares
Fermat's Factorization Method involves the concept of expressing a number as a difference of squares. This is based on the identity: \[ x^2 - y^2 = (x-y)(x+y) \]For an integer \(N\), the method seeks numbers \(x\) and \(y\) such that \(x^2 - y^2 = N\).
  • If these numbers are found, then \(N\) can be factored into \((x-y)(x+y)\).
The strategy involves selecting \(x\) such that \(x^2\) is slightly larger than \(N\). The goal is to make \(x^2-N\) a perfect square (which we'll discuss shortly). This is repeated progressively by adjusting \(x\) until we find integers \(x\) and \(y\) that satisfy \(x^2-N = y^2\).In practical application, such as factoring 2279, we begin close to the square root of 2279, incrementing if necessary, thus utilizing the difference of squares.Understanding this concept simplifies factorization compared to other complex methods, especially when numbers lack small factor pairs.
Perfect Square
A perfect square is simply a number that is the square of an integer. For example, 25 is a perfect square because it is \(5^2\).In Fermat's Factorization Method, once we have \(x^2 - N\), the important step is checking if this result is a perfect square, say \(y^2\).
  • If \(x^2 - N = y^2\), then we have successfully expressed \(N\) as a difference of squares.
  • This will allow us to write \(N = (x-y)(x+y)\).
In our example factorization of the number 2279, we found that \(48^2 - 2279 = 5^2\). Since 25 is a perfect square, we managed to express 2279 as the difference: \(48^2 - 5^2\).Recognizing perfect squares is crucial to using this method effectively, allowing for swift factorization once the right perfect square difference is encountered.
Integer Factorization
Integer factorization is the process of breaking down a composite number into a product of smaller integer factors. This process can be quite challenging for large numbers but is the ultimate aim when using Fermat's Factorization Method.This method seeks to simplify the search for factors by converting the problem into finding two perfect squares that differ by the given integer \(N\).
  • Such factorization is valuable in number theory and cryptography.
  • Especially for numbers where obvious prime factors aren't easily found.
Using Fermat's method, we effectively found that 2279 can be factored as \((48-5)(48+5) = 43 \times 53\). For practical problems, this method systematically reduces large integers into manageable factors without large divisibility tests.Ultimately, integer factorization reveals the inherent structure of the number, showing us how these separate squares come together to form the integer in question. Whether in textbooks or real-life applications like cryptography, factorization is a cornerstone of numerical analysis.

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Most popular questions from this chapter

Employ the generalized Fermat method to factor each of the following numbers: (a) \(2911\left[\right.\) Hint: \(138^{2}=67^{2}\) (mod 2911).] (b) 4573 [Hint: \(177^{2} \equiv 92^{2}\) (mod 4573).] (c) 6923 [Hint: \(208^{2} \equiv 93^{2}\) (mod 6923).]

Derive each of the following congruences: (a) \(a^{21} \equiv a(\bmod 15)\) for all \(a\). [Hint: By Fermat's theorem, \(\left.a^{5} \equiv a(\bmod 5) .\right]\) (b) \(a^{7}=a\) (mod 42) for all a. (c) \(a^{13} \equiv a(\bmod 3 \cdot 7 \cdot 13)\) for all \(a\). (d) \(a^{9} \equiv a\) (mod 30 ) for all \(a\).

(a) For a prime \(p\) of the form \(4 k+3\), prove that either $$ \left(\frac{p-1}{2}\right) ! \equiv 1(\bmod p) \quad \text { or } \quad\left(\frac{p-1}{2}\right) ! \equiv-1(\bmod p) $$ hence, \([(p-1) / 2] !\) satisfies the quadratic congruence \(x^{2} \equiv 1(\bmod p)\). (b) Use part (a) to show that if \(p=4 k+3\) is prime, then the product of all the even integers less than \(p\) is congruent modulo \(p\) to either 1 or \(-1\). [Hint: Fermat's theorem implies that \(\left.2^{(p-1) / 2} \equiv \pm 1(\bmod p) .\right]\)

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 \(x^{2}=(50-x)^{2}\) (mod 100\()\), it suffices to examine the final digits of \(x^{2}\) for the 26 values \(x=0,1,2, \ldots, 25\).]

Show that the smallest pseudoprime 341 is not an absolute pseudoprime by showing that \(11^{341} \neq 11(\bmod 341)\) [Hint: 31 \\} \(11^{341}-11 .\) ]
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