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April 21st, 2015, 10:48 AM  #1 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1  Primality test and factorization of LeporeSanto in logarithm
Primality test and factorization of LeporeSanto in log forgive translation errors Factorization and primality test of LeporeSanto in log [(YX) / 6] +1 I will show you the basic example that the factorization of two prime numbers because repeating the process you can factor any number. Each number NR (not multiple of 2 and 3)divided by six as decimal 1666p and 8333p (p stands for periodic) as per NR module six you will have 1/6= 0,1666p 2 is divisible by 2 3 is divisible by 3 4 is divisible by 2 5/6=0,8333p 6 is divisible by 2 and by 3 and this is equivalent to saying that, starting from 1 you should do once +4 once +2 , and then the set of numbers is not muliplo 2 and 3 will be: 1 1+4=5 5+2=7 7+4=11 11+2=13 13+4=17 17+2=19 19+4=23 23+2=25 25+4=29 29+2=31 â€¦.. â€¦.. ecc.ecc. then you can see this: 5*5; 25+(1*30) ; 25+(2*30); 25+(3*30); ecc. 5*7; 35+(1*30) ; 35+(2*30); 35+(3*30); ecc. 7*7; 49+(1*42); 49+(2*42); ecc. ecc. 7*11; 77+(1*42); 77+(2*42); ecc. ecc. From this we derive the three equations: X^2+n*(X*6)=NR X*(X+2)+n(X*6)=NR X*(X+4)+n(X*6)=NR where n=(YX)/6 in the first equation n=(YX2)/6 in the second equation n=(YX4)/6 in the third equation So if we find the value of n we find X. Alberico Lepore 12/04/2015 The idea is to divide by 2, n, until you get to n = 1 (test for n = 1 if the equation transformed from results). If n is even okay, if n is odd we must remove one and divide by two. How the algorithm works. Consider 6001=17*357 since 6001/6=1000,1666p and therefore has as a decimal part 1666p we will use the first equation: X^2+n*(X*6)=6001 and n=56 then 56/2=28 28/2=14 14/2=7 then 71=6 6/2=3 then 31=2 2/1=1 then the sequence is par par par odd odd So we will modify X^2+n*(X*6)=6001 in to X^2+n*(X*12)=6001 X^2+n*(X*24)=6001 X^2+n*(X*4=6001 We are in the odd then remove 1 to n which means add X * 48 and divide by 2, and then X^2+n*(X*96)+48X=6001 yet so odd X^2+n*(X*192)+96X+48X=6001 then X^2+n*(X*192)+144X=6001 for n=1 X^2+(X*336)=6001 you will X=17 Now let's see how do I decide whether n is par or is odd we bring the equation in the form n*X=(6001 â€“ X^2)/6 The left side equals the right side so if a is par is par to the other also, if it is a odd is odd the other also. We must consider that X is always odd, then n*X will be par if n is par and will be odd if n is odd. We see an example of the procedure. 6001/6=1000,166p (X^2)/6=k,1666 because the squares are always in this form So we remove 0,1666p to 1000,1666p and we take K,1666p will become k then we would have 1000k which is an integer and is to be evaluated if it is par or is odd we must find the integer values of n The equation of departure x^2 + n*(6x) = 6001 k is always par because x^2=(6k+5)^2=36k^2+60k+25 or x^2=(6k+1)^2=36k^2+12k+1 both if you take away one and divide by 6 are par then 1000k=par then n par So the new equation will be x ^ 2 + n * (12x) = 6001 we divide by 2 since k is par. k/2 is an integer then the odd or par value of n will be given only 1000/2 = 500 which is an integer then n is par (as we already knew). 500k / 2 = integer see now n / 2 is? 500/2 = 250 we want only the integer values ??of n therefore 250k/4=integer then 500k2 = par then that n / 2 is par and then also that k / 2 is par then we did well see now n / 4 is? 250 / 2 = 125 we want only the integer values ??of n therefore 125k / 8 = integer, then 250k/4=par then that n / 4 is par and then also that k / 4 is par then we did well see now n / 8 is? 125/2=62,5 we want only the integer values of n, hence 62.5k / 16 = integer, then 125k / 8 = ODD follows that n / 8 is ODD and therefore also that k / 8 is not an integer then we go wrong because k / 4 was par we leave unchanged n and integers and take 62. 62k / 16 = integer then k / 8 = par therefore would be fine for k / 4 62k / 16 = integer then that 125k / 8 = odd Now we can say that n / 8 is odd Since n / 8 is odd I must subtract 1 and divide by two. Subtract means subtract 1 to 62 then become 61 see now n / 16 is? 61/2 = 30.5 we want only the integer values of n, hence 30.5k / 32 = integer, then 61k/16 = odd so that n / 16 is ODD why this is true? because k/8 is par and n/8 Ã¨ odd. in the meantime, we have transformed the equation (and at each step we tried for n = 1) that has become X ^ 2 + n * (X * 192) + 144x = 6001 for n = 1 X ^ 2 + (X * 336) = 6001 you will have X = 17 at this time we are aware of 56 because coming from below and knowing the sequence of odd and par we can get, but we do not care to know PPPDD is our sequence ATTENTION MUST ALWAYS KEEP the K PAR BEFORE THE DIVISION Alberico Lepore, Ruggiero Santo 15/04/2015 For cases NR / 6 = k, 83333 is a bit different: You multiply the square number to ensure that it becomes a (6 * numeronaturale) +1 that is my k, 16666p, then the new equation will be: Z^2+n*(Z*6)=NR where Z=X^2 Z is proceeding normally, and you to find the square root of X. The time taken will log{[(YX)^2]/6}+1 Alberico Lepore, Ruggiero Santo 20/04/2015 clearly it is here, in Italian Test di primalitÃ e fattorizzazione di LeporeSanto in log  How to decode RSA we would like your opinion 
April 21st, 2015, 11:09 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I'm having trouble following the language. Do you have any code available? that might be easier to understand.

April 21st, 2015, 12:00 PM  #3 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
unfortunately I am not a mathematician and not a computer

April 22nd, 2015, 05:24 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
April 22nd, 2015, 05:29 AM  #5  
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1  Quote:
You will know the end. but do not need to know.  
April 22nd, 2015, 05:31 AM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
So it's not really a factorization method, then, just a certificate of compositeness.

April 22nd, 2015, 05:37 AM  #7  
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
He is a primality test and factorization We are interested to know the sequence of par and odd. Quote:
 
April 22nd, 2015, 07:24 AM  #8 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
Sorry there was an error. Now is good. Now let's see how do I decide whether n is par or is odd we bring the equation in the form n*X=(6001 â€“ X^2)/6 The left side equals the right side so if a is par is par to the other also, if it is a odd is odd the other also. We must consider that X is always odd, then n*X will be par if n is par and will be odd if n is odd. We see an example of the procedure. 6001/6=1000,166p (X^2)/6=k,1666 because the squares are always in this form So we remove 0,1666p to 1000,1666p and we take K,1666p will become k then we would have 1000k which is an integer and is to be evaluated if it is par or is odd we must find the integer values of n The equation of departure x^2 + n*(6x) = 6001 k is always par because x^2=(6k+5)^2=36k^2+60k+25 or x^2=(6k+1)^2=36k^2+12k+1 both if you take away one and divide by 6 are par then 1000k=par then n par So the new equation will be x ^ 2 + n * (12x) = 6001 we divide by 2 since k is par. k/2 is an integer then the odd or par value of n will be given only 1000/2 = 500 which is an integer then n is par (as we already knew). 500k / 2 = integer see now n / 2 is? 500/2 = 250 we want only the integer values of k therefore 250k/4=integer then 500k2 = par then that n / 2 is par see now n / 4 is? 250 / 2 = 125 we want only the integer values of k therefore 125k / 8 = integer, then 250k/4=par then that n / 4 is par see now n / 8 is? 125/2=62,5 we want only the integer values of k, hence 62.5k / 16 = integer, then 125k / 8 = ODD follows that n / 8 is ODD Now we can say that n / 8 is odd Since n / 8 is odd I must subtract 1 and divide by two. Subtract means subtract 1 to 62 then become 61 see now n / 16 is? 61/2 = 30.5 we want only the integer values of k, hence 30.5k / 32 = integer, then 61k/16 = odd so that n / 16 is ODD in the meantime, we have transformed the equation (and at each step we tried for n = 1) that has become X ^ 2 + n * (X * 192) + 144x = 6001 for n = 1 X ^ 2 + (X * 336) = 6001 you will have X = 17 at this time we are aware of 56 because coming from below and knowing the sequence of odd and par we can get, but we do not care to know PPPDD is our sequence 
April 22nd, 2015, 10:43 AM  #9 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
This is correct part () Factorization and primality test of LeporeSanto in log [(YX)/6]+1 I will show you the basic example, that i sto say the factorization of two prime numbers because reterating this process we can factor any number. Each number NR (not multiple of 2 and 3) divided by six gives as decimal 1666p and 8333p (the p is periodic) as for each NR module six we will have 1/6= 0,1666p 2 is divisible by 2 3 is divisible by 3 4 is divisible by 2 5/6=0,8333p 6 is divisible by 2and 3 and this is the same if we start from 1 , doing before +4 and then +2 , so the set of numbers not multiple of 2 and 3 will be: 1 1+4=5 5+2=7 7+4=11 11+2=13 13+4=17 17+2=19 19+4=23 23+2=25 25+4=29 29+2=31 â€¦.. â€¦.. ecc.ecc. So we can see this: 5*5; 25+(1*30) ; 25+(2*30); 25+(3*30); ecc. 5*7; 35+(1*30) ; 35+(2*30); 35+(3*30); ecc. 7*7; 49+(1*42); 49+(2*42); ecc. ecc. 7*11; 77+(1*42); 77+(2*42); ecc. ecc. From this we can derive the three equations: X^2+n*(X*6)=NR X*(X+2)+n(X*6)=NR X*(X+4)+n(X*6)=NR n=(YX)/6 in the first equation n=(YX2)/6 in the second equation n=(YX4)/6 in the third equation So if we find the value of n we can get X. Alberico Lepore 12/04/2015 The idea is to divide by 2 the n until to come at n = 1 (so test for n = 1 if the equation transformed produced results). If n is even itâ€™s okay, but if n is odd we must remove 1 and divide by two. How the algorithm works. Consider 6001=17*357 given that 6001/6=1000,1666p and therefore has as part decimal 1666p we will use the first equation, in fact: X ^ 2 + n * (X * 6) = 6001 and n = 56 Donâ€™t worry if you donâ€™t understand how I know if n is equal or odd Iâ€™ll tell you later. then 56/2 = 28 2/28 = 14 14/2 = 7 so 71 = 6 6/2 = 3 so 31 = 2 2/1 = 1 So the sequence is Even E E Odd O Then we can modify X ^ 2 + n * (X * 6) = 6001 to X ^ 2 + n * (X * 12) = 6001 X ^ 2 + n * (X * 24) = 6001 X ^ 2 + n * (X * 4 = 6001 We are in the odd so remove 1 to n that means adding X*48 more divide by 2: X^2+n*(X*96)+48X=6001 yet odd so X^2+n*(X*192)+96X+48X=6001 that better written is X^2+n*(X*192)+144X=6001 for n = 1 X^2+(X*336)=6001 we will have X = 17 Now letâ€™s see how do to decide if n is even or odd we bring the equation in the form n*X=(6001 â€“ X^2)/6 The left part is equal to the right part so if one is equal , is equal also the other ; if one is odd, is odd the other also. We have to consider that X is always odd then n * X will be even if n is even and will be odd if n is odd. We see the procedure with an example 6001/6=1000,166p (X^2)/6=k,1666 because squares are always in this form So we remove 0,1666p to 1000,1666p and we remove to K, 1666p that will become k then we would have 1000k that is an integer and we need to evaluate if it is even or is odd we have to find the integer values of n. So the equation x^2 + n*(6x) = 6001 k is always even because x^2=(6k+5)^2=36k^2+60k+25 or x^2=(6k+1)^2=36k^2+12k+1 if you take away one and divide by 6 both are equal it follows. So 1000k is even and then n is even So the new equation will be x ^ 2 + n * (12x) = 6001 , we divide by 2 TO BE CONTINUED THIS PART it is missing hereâ€¦â€¦â€¦â€¦. â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€ ”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â €”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€” other OTHER PART For the case NR/6=k,83333 is different: We multiply square number to ensure that it becomes a (6 * natural number) +1 so is my k,16666p , the new equation will be: Z^2+n*(Z*6)=NR Where Z=X^2 so is normally find Z and we do square root to find X. The time used will be log {[(YX) ^ 2] / 6} +1 Alberico Lepore, Ruggiero Santo 22/04/2015 clearly it is here, in Italian Test di primalitÃ e fattorizzazione di LeporeSanto in log  How to decode RSA 
April 22nd, 2015, 10:04 PM  #10 
Member Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 
the missing part is momentarily in Italian. Waiting for translation Factorization and primality test of LeporeSanto in log [(YX)/6]+1 I will show you the basic example, that i sto say the factorization of two prime numbers because reterating this process we can factor any number. Each number NR (not multiple of 2 and 3) divided by six gives as decimal 1666p and 8333p (the p is periodic) as for each NR module six we will have 1/6= 0,1666p 2 is divisible by 2 3 is divisible by 3 4 is divisible by 2 5/6=0,8333p 6 is divisible by 2and 3 and this is the same if we start from 1 , doing before +4 and then +2 , so the set of numbers not multiple of 2 and 3 will be: 1 1+4=5 5+2=7 7+4=11 11+2=13 13+4=17 17+2=19 19+4=23 23+2=25 25+4=29 29+2=31 â€¦.. â€¦.. ecc.ecc. So we can see this: 5*5; 25+(1*30) ; 25+(2*30); 25+(3*30); ecc. 5*7; 35+(1*30) ; 35+(2*30); 35+(3*30); ecc. 7*7; 49+(1*42); 49+(2*42); ecc. ecc. 7*11; 77+(1*42); 77+(2*42); ecc. ecc. From this we can derive the three equations: X^2+n*(X*6)=NR X*(X+2)+n(X*6)=NR X*(X+4)+n(X*6)=NR n=(YX)/6 in the first equation n=(YX2)/6 in the second equation n=(YX4)/6 in the third equation So if we find the value of n we can get X. Alberico Lepore 12/04/2015 The idea is to divide by 2 the n until to come at n = 1 (so test for n = 1 if the equation transformed produced results). If n is even itâ€™s okay, but if n is odd we must remove 1 and divide by two. How the algorithm works. Consider 6001=17*357 given that 6001/6=1000,1666p and therefore has as part decimal 1666p we will use the first equation, in fact: X ^ 2 + n * (X * 6) = 6001 and n = 56 Donâ€™t worry if you donâ€™t understand how I know if n is equal or odd Iâ€™ll tell you later. then 56/2 = 28 2/28 = 14 14/2 = 7 so 71 = 6 6/2 = 3 so 31 = 2 2/1 = 1 So the sequence is Even E E Odd O Then we can modify X ^ 2 + n * (X * 6) = 6001 to X ^ 2 + n * (X * 12) = 6001 X ^ 2 + n * (X * 24) = 6001 X ^ 2 + n * (X * 4 = 6001 We are in the odd so remove 1 to n that means adding X*48 more divide by 2: X^2+n*(X*96)+48X=6001 yet odd so X^2+n*(X*192)+96X+48X=6001 that better written is X^2+n*(X*192)+144X=6001 for n = 1 X^2+(X*336)=6001 we will have X = 17 Now letâ€™s see how do to decide if n is even or odd we bring the equation in the form n*X=(6001 â€“ X^2)/6 The left part is equal to the right part so if one is equal , is equal also the other ; if one is odd, is odd the other also. We have to consider that X is always odd then n * X will be even if n is even and will be odd if n is odd. We see the procedure with an example 6001/6=1000,166p (X^2)/6=k,1666 because squares are always in this form So we remove 0,1666p to 1000,1666p and we remove to K, 1666p that will become k then we would have 1000k that is an integer and we need to evaluate if it is even or is odd we have to find the integer values of n. So the equation x^2 + n*(6x) = 6001 k is always even because x^2=(6h+5)^2=36h^2+60h+25 or x^2=(6h+1)^2=36h^2+12h+1 if you take away one and divide by 6 both are equal it follows. So 1000k is even and then n is even So the new equation will be x ^ 2 + n * (12x) = 6001 k/2 Ã¨ dispari se n=6H+5 k/2 Ã¨ dispari se n=6H+1 k/2 Ã¨ dispari n=3F con F dispari n =1000k 1000/6=166,6666 166666*6=996 n =1000k^2=996[k4] 996/6=166 e [k4] Ã¨ pari quindi non puÃ² essere dispari nei primi due casi vediamo il terzo 1000/3=333,333 n =3*(333)k=3*(333)[k1] 333 Ã¨ dispari e [(k1]=DISPARI significa che anche se Ã¨ diviso 3, F Ã¨ pari quindi k/2 Ã¨ pari ora possiamo dividiamo per 2 quindi n/2=500 â€“ k/2=pari velocizzando un po si ha n/2=498((k/2)2) 498/6=83 e ((k/2)2) Ã¨ pari quindi non puÃ² essere dispari nei primi due casi (498/3)=166 Ã¨ pari e ((k/2)2)=pari significa che anche se Ã¨ diviso 3, F Ã¨ pari k/4=pari ora possiamo dividiamo per 2 quindi n/4=250 â€“ k/4=pari n/4=246[(k/4)4] 246/6=246 e [(k/4)4] Ã¨ pari quindi non puÃ² essere dispari nei primi due casi n/4=249[(k/4)1] (249/3)=83 Ã¨ dispari e [(k/4)1] Ã¨ dispari significa che anche se Ã¨ diviso 3, F Ã¨ pari k/8=pari ora possiamo dividiamo per 2 quindi n/8=125 â€“ k/8=dispari siccome Ã¨ dispari dobbiamo togliere 1 a 125 che diventerÃ 124 n/8=120[(k / 8 )5] 120/6=20 e [(k / 8 )5] quindi n/8=(6*20) [(k / 8 )5] siccome k/8 Ã¨ pari Ã¨ divisibile per 2 quindi dobbiamo vedere se k/8 Ã¨ divisibile anche per 3 ma ricordiamo che k Ã¨ sempre divisibile per tre perchÃ¨ (x^2)1=(6h+5)^2=36h^2+60h+251 oppure (x^2)1=(6h+1)^2=36h^2+12h+1 di conseguenza Ã¨ divisibile per 2 e per tre quindi Ã¨ divisibile per 6 quindi n/8=6*(20G)+5 dove G= k/4 quindi k/16 Ã¨ dispari ora possiamo dividiamo per 2 quindi n/16=62 â€“ k/16=dispari Alberico Lepore, Ruggiero Santo 23/04/2015 â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€ ”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â €”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€” â€”â€” for cases NR/6=k,83333 behave similarly Alberico Lepore, Ruggiero Santo 23/04/2015 Last edited by gerva; April 22nd, 2015 at 10:58 PM. 

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