factorization methods
play

Factorization Methods Bernd Schr oder Bernd Schr oder Louisiana - PowerPoint PPT Presentation

Nov 04, 2023 •510 likes •2.03k views

Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Factorization Methods Bernd Schr oder Bernd Schr oder Louisiana Tech University, College of Engineering and Science Factorization Methods Trial Division

  1. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let n be an odd positive integer. There is a one-to-one correspondence between the factorizations of n into two positive integers and differences of two squares that equal n. Proof. To write a factorization of n as ab = n = t 2 − s 2 , use t = a + b 2 , s = a − b 2 . The one-to-one correspondence (injective function) is that we consider the factorization itself as input ( a , b ) and f ( a , b ) = ( t , s ) Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  2. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let n be an odd positive integer. There is a one-to-one correspondence between the factorizations of n into two positive integers and differences of two squares that equal n. Proof. To write a factorization of n as ab = n = t 2 − s 2 , use t = a + b 2 , s = a − b 2 . The one-to-one correspondence (injective function) is that we consider the factorization itself as input � a + b 2 , a − b � ( a , b ) and f ( a , b ) = ( t , s ) = . 2 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  3. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let n be an odd positive integer. There is a one-to-one correspondence between the factorizations of n into two positive integers and differences of two squares that equal n. Proof. To write a factorization of n as ab = n = t 2 − s 2 , use t = a + b 2 , s = a − b 2 . The one-to-one correspondence (injective function) is that we consider the factorization itself as input � a + b 2 , a − b � ( a , b ) and f ( a , b ) = ( t , s ) = . 2 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  4. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  5. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  6. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  7. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Then s 2 = ( t + k ) 2 − n Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  8. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Then s 2 = ( t + k ) 2 − n and n = ( t + k ) 2 − s 2 = ( t + k + s )( t + k − s ) Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  9. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Then s 2 = ( t + k ) 2 − n and n = ( t + k ) 2 − s 2 = ( t + k + s )( t + k − s ) � 2 − � 2 we will find a number that � n + 1 � n − 1 3. Because n = 2 −√ n steps. 2 2 works in at most n Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  10. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Then s 2 = ( t + k ) 2 − n and n = ( t + k ) 2 − s 2 = ( t + k + s )( t + k − s ) � 2 − � 2 we will find a number that � n + 1 � n − 1 3. Because n = 2 −√ n steps. 2 2 works in at most n 4. But we cannot jump directly to the end Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  11. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Implementation 1. Take t to be the smallest integer greater than √ n . 2. Look for perfect squares s 2 in the sequence t 2 − n , ( t + 1 ) 2 − n , ( t + 2 ) 2 − n , ... . Then s 2 = ( t + k ) 2 − n and n = ( t + k ) 2 − s 2 = ( t + k + s )( t + k − s ) � 2 − � 2 we will find a number that � n + 1 � n − 1 3. Because n = 2 −√ n steps. 2 2 works in at most n 4. But we cannot jump directly to the end, because using the above difference only leads to the factorization n = n · 1. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  12. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  13. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. ◮ Factoring 1 , 363 takes one step with this method. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  14. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. ◮ Factoring 1 , 363 takes one step with this method. ◮ Factoring 1 , 463 takes 10 steps. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  15. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. ◮ Factoring 1 , 363 takes one step with this method. ◮ Factoring 1 , 463 takes 10 steps. ◮ Unfortunately, for products of factors that are not approximately equal, Fermat factorization can take a lot longer: Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  16. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. ◮ Factoring 1 , 363 takes one step with this method. ◮ Factoring 1 , 463 takes 10 steps. ◮ Unfortunately, for products of factors that are not approximately equal, Fermat factorization can take a lot longer: 1 , 461 is divisible by 3 and Fermat factorization does not yield a factorization in 100 steps. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  17. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Example. ◮ Factoring 1 , 363 takes one step with this method. ◮ Factoring 1 , 463 takes 10 steps. ◮ Unfortunately, for products of factors that are not approximately equal, Fermat factorization can take a lot longer: 1 , 461 is divisible by 3 and Fermat factorization does not yield a factorization in 100 steps. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  18. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  19. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  20. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  21. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  22. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  23. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  24. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  25. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  26. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 641 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  27. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 641 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  28. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  29. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  30. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  31. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  32. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  33. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 641 · 2 28 − � = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  34. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 = 641 · 2 28 − ( 641 − 1 ) 4 + 1 641 · 2 28 − � = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  35. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 = 641 · 2 28 − ( 641 − 1 ) 4 + 1 641 · 2 28 − � = � 2 28 − 641 3 + 4 · 641 2 − 6 · 641 + 4 � = 641 · Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  36. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 = 641 · 2 28 − ( 641 − 1 ) 4 + 1 641 · 2 28 − � = � 2 28 − 641 3 + 4 · 641 2 − 6 · 641 + 4 � = 641 · So 641 | 2 2 5 + 1. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  37. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 = 641 · 2 28 − ( 641 − 1 ) 4 + 1 641 · 2 28 − � = � 2 28 − 641 3 + 4 · 641 2 − 6 · 641 + 4 � = 641 · So 641 | 2 2 5 + 1. Now do the division Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  38. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Numbers of the form 2 2 n + 1 are called Fermat numbers . Fermat conjectured that all these numbers are prime. This conjecture turned out to be incorrect. Example. 2 2 5 + 1 = 641 · 6 , 700 , 417 (Euler, 1732). Proof. 5 · 2 7 + 1 = 2 4 + 5 4 = 641 2 2 5 + 1 2 32 + 1 = 2 4 · 2 28 + 1 = � 641 − 5 4 � 2 28 + 1 = 5 · 2 7 � 4 + 1 = 641 · 2 28 − ( 641 − 1 ) 4 + 1 641 · 2 28 − � = � 2 28 − 641 3 + 4 · 641 2 − 6 · 641 + 4 � = 641 · So 641 | 2 2 5 + 1. Now do the division Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  39. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  40. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  41. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  42. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  43. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  44. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  45. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  46. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  47. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  48. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  49. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Example. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  50. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Example. 2 2 6 + 1 is not prime Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  51. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Example. 2 2 6 + 1 is not prime, but it takes a while until a factor 256 · 1071 + 1 is found. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  52. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations So the next question is which Fermat numbers are prime. Theorem (proof is later an exercise). Every prime divisor of a Fermat number 2 2 n + 1 is of the form 2 n + 1 k + 1 Example. 2 2 3 + 1 is prime, because 2 4 k + 1 = 16 k + 1 and 2 2 3 + 1 = 257, which is not divisible by 17 and all other numbers of the form 16 k + 1 are too large to be factors, actually, 17 is, too. Example. 2 2 6 + 1 is not prime, but it takes a while until a factor 256 · 1071 + 1 is found. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  53. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  54. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers 1. Only the first four Fermat numbers are known to be prime: 2 2 1 + 1 = 5, 2 2 2 + 1 = 17, 2 2 3 + 1 = 257, 2 2 4 + 1 = 65 , 537. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  55. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers 1. Only the first four Fermat numbers are known to be prime: 2 2 1 + 1 = 5, 2 2 2 + 1 = 17, 2 2 3 + 1 = 257, 2 2 4 + 1 = 65 , 537. 2. Another 243 are known to be composite. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  56. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers 1. Only the first four Fermat numbers are known to be prime: 2 2 1 + 1 = 5, 2 2 2 + 1 = 17, 2 2 3 + 1 = 257, 2 2 4 + 1 = 65 , 537. 2. Another 243 are known to be composite. 3. So now there also is a conjecture that only the first four Fermat numbers are prime. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  57. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers 1. Only the first four Fermat numbers are known to be prime: 2 2 1 + 1 = 5, 2 2 2 + 1 = 17, 2 2 3 + 1 = 257, 2 2 4 + 1 = 65 , 537. 2. Another 243 are known to be composite. 3. So now there also is a conjecture that only the first four Fermat numbers are prime. 4. The research in this direction certainly tests the limits of what is possible: There are only 7 composite Fermat numbers for which we know the complete factorization. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  58. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Facts About Fermat Numbers 1. Only the first four Fermat numbers are known to be prime: 2 2 1 + 1 = 5, 2 2 2 + 1 = 17, 2 2 3 + 1 = 257, 2 2 4 + 1 = 65 , 537. 2. Another 243 are known to be composite. 3. So now there also is a conjecture that only the first four Fermat numbers are prime. 4. The research in this direction certainly tests the limits of what is possible: There are only 7 composite Fermat numbers for which we know the complete factorization. (Think about it: 2 2 12 + 1 = 2 4096 + 1 > 10 1365 .) Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  59. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  60. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  61. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  62. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  63. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  64. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  65. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  66. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  67. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  68. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  69. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  70. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  71. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n = ( F n − 2 ) F n Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  72. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n = ( F n − 2 ) F n 2 2 n − 1 2 2 n + 1 � �� � = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  73. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n = ( F n − 2 ) F n 2 2 n − 1 2 2 n + 1 � �� � = 2 2 n + 2 n − 1 � � = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  74. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n = ( F n − 2 ) F n 2 2 n − 1 2 2 n + 1 � �� � = 2 2 n + 2 n − 1 � � = 2 2 n + 1 + 1 − 2 � � = Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

  75. Trial Division Fermat Factorization Fermat Numbers Linear Diophantine Equations Lemma. Let F k : = 2 2 k + 1 . Then for all positive integers n we have F 0 F 1 F 2 ··· F n − 1 = F n − 2 . Proof. Induction on n . Base Step, n = 2 . F 0 F 1 = 3 · 5 = 15 = 17 − 2 = F 2 − 2 . Induction Step, n → n + 1 . F 0 F 1 F 2 ··· F n = ( F 0 F 1 F 2 ··· F n − 1 ) F n = ( F n − 2 ) F n 2 2 n − 1 2 2 n + 1 � �� � = 2 2 n + 2 n − 1 � � = 2 2 n + 1 + 1 − 2 � � = = F n + 1 − 2 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Factorization Methods

Recommend

Integer Factorization Methods Modular Arithmetic Trial division, Pollards p 1 , Division

Integer Factorization Methods Modular Arithmetic Trial division, Pollards p 1 , Division

Integer Factorization Methods Integer 2014-04-11 Integer Factorization Methods Factorization Trial division, Pollards p 1 , Pollards , and Fermats method Methods Christopher Koch 1 C. Koch 1 Department of Computer Science and

681 views • 33 slides
Integer Factorization Methods Modular Trial division, Pollards p 1 , Arithmetic Division

Integer Factorization Methods Modular Trial division, Pollards p 1 , Arithmetic Division

Integer Factorization Methods C. Koch Overview Integer Factorization Methods Modular Trial division, Pollards p 1 , Arithmetic Division Algorithm and Congruence Pollards , and Fermats method Residue classes mod n Integers

505 views • 33 slides
Fast Coordinate Descent methods for Non-Negative Matrix Factorization Inderjit S. Dhillon

Fast Coordinate Descent methods for Non-Negative Matrix Factorization Inderjit S. Dhillon

Non-negative Matrix Factorization Fast Coordinate Descent methods for Non-Negative Matrix Factorization Inderjit S. Dhillon University of Texas at Austin SIAM Conference on Applied Linear Algebra Valencia, Spain June 19, 2012 Joint work with

682 views • 25 slides
Structured sparse methods for matrix factorization Francis Bach Sierra team, INRIA - Ecole

Structured sparse methods for matrix factorization Francis Bach Sierra team, INRIA - Ecole

Structured sparse methods for matrix factorization Francis Bach Sierra team, INRIA - Ecole Normale Sup erieure March 2011 Joint work with R. Jenatton, J. Mairal, G. Obozinski Structured sparse methods for matrix factorization Outline

1.07k views • 73 slides
Matrix Factorization and Factorization Machines for Recommender Systems Chih-Jen Lin Department

Matrix Factorization and Factorization Machines for Recommender Systems Chih-Jen Lin Department

Matrix Factorization and Factorization Machines for Recommender Systems Chih-Jen Lin Department of Computer Science National Taiwan University Talk at SDM workshop on Machine Learning Methods on Recommender Systems, May 2, 2015 Chih-Jen Lin

1.14k views • 54 slides
Chapter IX: Matrix factorizations* 1. The general idea 2. Matrix factorization methods 3. Latent

Chapter IX: Matrix factorizations* 1. The general idea 2. Matrix factorization methods 3. Latent

Chapter IX: Matrix factorizations* 1. The general idea 2. Matrix factorization methods 3. Latent topic models 4. Dimensionality reduction *Zaki & Meira, Ch. 8; Tan, Steinbach & Kumar, App. B; Manning, Raghavan & Schtze, Ch. 18

798 views • 53 slides
Direct Methods for Solving Linear Systems Matrix Factorization Numerical Analysis (9th Edition)

Direct Methods for Solving Linear Systems Matrix Factorization Numerical Analysis (9th Edition)

Direct Methods for Solving Linear Systems Matrix Factorization Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University 2011 Brooks/Cole, Cengage Learning c

1.5k views • 132 slides
Parallel Numerical Algorithms Chapter 4 Sparse Linear Systems Section 4.1 Direct Methods

Parallel Numerical Algorithms Chapter 4 Sparse Linear Systems Section 4.1 Direct Methods

Sparse Matrices Sparse Triangular Solve Cholesky Factorization Sparse Cholesky Factorization Parallel Numerical Algorithms Chapter 4 Sparse Linear Systems Section 4.1 Direct Methods Michael T. Heath and Edgar Solomonik Department of

906 views • 63 slides
Compressed Factorization: Fast and Accurate Low-Rank Factorization of Compressively-Sensed Data

Compressed Factorization: Fast and Accurate Low-Rank Factorization of Compressively-Sensed Data

Compressed Factorization: Fast and Accurate Low-Rank Factorization of Compressively-Sensed Data Vatsal Sharan* , Kai Sheng Tai*, Peter Bailis & Gregory Valiant Stanford University Poster 187 Learning(from(compressed(data

253 views • 8 slides
Factoring Done by:Rashed salmeen Grade:9ASP2 Prime factorization Prime factorization:is finding

Factoring Done by:Rashed salmeen Grade:9ASP2 Prime factorization Prime factorization:is finding

Factoring Done by:Rashed salmeen Grade:9ASP2 Prime factorization Prime factorization:is finding which prime numbers multiply together to make the original number. Examples: 27 18 2 9 3 9 3 3 3 3 18=2x3x3 27=3x3x3 greatest common

365 views • 11 slides
AMath 483/583 Lecture 23 Notes: Outline: Linear systems: LU factorization and condition

AMath 483/583 Lecture 23 Notes: Outline: Linear systems: LU factorization and condition

AMath 483/583 Lecture 23 Notes: Outline: Linear systems: LU factorization and condition number Heat equation and discretization Iterative methods Sample codes: $UWHPSC/codes/openmp/jacobi1d_omp1.f90

255 views • 10 slides
L101: Matrix Factorization In a nutshell Matrix factorization/completion you know? In NLP?

L101: Matrix Factorization In a nutshell Matrix factorization/completion you know? In NLP?

L101: Matrix Factorization In a nutshell Matrix factorization/completion you know? In NLP? Word embeddings Topic models Information extraction FastText Why complete the matrix? Label Features Label Label Label Label

420 views • 20 slides
Incomplete Factorization by Local Exact Factorization (ILUE) Johannes Kraus and Maria Lymbery

Incomplete Factorization by Local Exact Factorization (ILUE) Johannes Kraus and Maria Lymbery

Outline The ILUE preconditioner Application within ASMG preconditioning Numerical results Incomplete Factorization by Local Exact Factorization (ILUE) Johannes Kraus and Maria Lymbery "Modelling 2014" June 26, 2014, Roznov Pod

771 views • 43 slides
Online-Updating Regularized Kernel Matrix Factorization Models for Large-Scale Recommender

Online-Updating Regularized Kernel Matrix Factorization Models for Large-Scale Recommender

Introduction Matrix factorization Learning Matrix Factorization Models Online Updates Evaluation Online-Updating Regularized Kernel Matrix Factorization Models for Large-Scale Recommender Systems. Steffen Rendle, Lars Schmidt-Thieme

903 views • 21 slides
LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University

LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University

DM559 Linear and Integer Programming LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark [ Based on slides by Lieven Vandenberghe, UCLA ] Operation Count LU Factorization Outline

935 views • 34 slides
LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University

LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University

DM559 Linear and Integer Programming LU Factorization Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark [ Based on slides by Lieven Vandenberghe, UCLA ] Operation Count LU Factorization Outline

524 views • 38 slides
Quadratic Sieve implementation for factorization backup, benchmark and network communication

Quadratic Sieve implementation for factorization backup, benchmark and network communication

Quadratic Sieve implementation for factorization backup, benchmark and network communication Ayoub Ouarrak 26/04/2016 Universita degli studi di Parma Dipartimento di Matematica e Informatica Introduction RSA and factorization RSA is a

833 views • 36 slides
Factorization on GPUs Wei Tan IBM T. J. Watson Research Center wtan@us.ibm.com

Factorization on GPUs Wei Tan IBM T. J. Watson Research Center wtan@us.ibm.com

cuMF_sgd: Fast and Scalable Matrix Factorization on GPUs Wei Tan IBM T. J. Watson Research Center wtan@us.ibm.com http://github.com/cumf http://researcher.ibm.com/person/us-wtan GTC 2017 Agenda Why accelerate matrix factorization?

571 views • 21 slides
Random problems at the core of integer factorization algorithms Pierrick Gaudry Caramba

Random problems at the core of integer factorization algorithms Pierrick Gaudry Caramba

Random problems at the core of integer factorization algorithms Pierrick Gaudry Caramba LORIA, Nancy CNRS, Universit de Lorraine, Inria Journes ALEA, March 2016 1/31 Plan Introduction: crypto context Integer factorization 101 How

462 views • 33 slides
The Scalable Petascale Data-Driven Approach for the Cholesky Factorization with multiple GPUs

The Scalable Petascale Data-Driven Approach for the Cholesky Factorization with multiple GPUs

The Scalable Petascale Data-Driven Approach for the Cholesky Factorization with multiple GPUs Yuki Tsujita, Toshio Endo, Katsuki Fujisawa Tokyo Institute of Technology, Japan ESPM2 2015@Austin Texas, USA 1 What is the Cholesky factorization?

743 views • 25 slides
TIME-DEPENDENT PARAMETRIC AND HARMONIC TEMPLATES IN NON-NEGATIVE MATRIX FACTORIZATION 13 th

TIME-DEPENDENT PARAMETRIC AND HARMONIC TEMPLATES IN NON-NEGATIVE MATRIX FACTORIZATION 13 th

Introduction Non negative-Matrix Factorization Spectrogram model Decomposition of musical spectrograms Conclusion TIME-DEPENDENT PARAMETRIC AND HARMONIC TEMPLATES IN NON-NEGATIVE MATRIX FACTORIZATION 13 th International Conference on Digital

750 views • 26 slides
Matrix Factorization and Lifting Palle Jorgensen and Myung-Sin Song The University of Iowa,

Matrix Factorization and Lifting Palle Jorgensen and Myung-Sin Song The University of Iowa,

Matrix Factorization and Lifting Palle Jorgensen and Myung-Sin Song The University of Iowa, Southern Illinois University Edwardsville January 7, 2011 Introduction Develop and refine a procedure which uses factorization of families of

544 views • 35 slides
Tensor Factorization via Matrix Factorization Volodymyr Kuleshov Arun Tejasvi Chaganty Percy

Tensor Factorization via Matrix Factorization Volodymyr Kuleshov Arun Tejasvi Chaganty Percy

Tensor Factorization via Matrix Factorization Volodymyr Kuleshov Arun Tejasvi Chaganty Percy Liang Stanford University May 11, 2015 Kuleshov, Chaganty, Liang (Stanford University) Tensor Factorization May 11, 2015 1 / 28 Introduction:

1.1k views • 79 slides
A Model For Mixed Linear-Tropical Matrix Factorization James Hook, Sanjar Karaev, Pauli Miettinen

A Model For Mixed Linear-Tropical Matrix Factorization James Hook, Sanjar Karaev, Pauli Miettinen

A Model For Mixed Linear-Tropical Matrix Factorization James Hook, Sanjar Karaev, Pauli Miettinen University of Birmingham: 18th June 2018 Low-Rank Approximate Factorization Given a matrix A R n m , an approximate factorization of rank k

207 views • 17 slides

More recommend