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