Quoting Marc Branchaud <[EMAIL PROTECTED]>: > > Anyone know if there's any truth to this? > > Marc > > > MANILA BULLETIN: http://www.mb.com.ph/INFO/2001-02/IT020201.asp > > Pinoy math enthusiast finds fast way to decode RSA encryption > At 03-02-01 02:27, Marc Branchaud wrote: After finding his e-mail adres, I just asked him : From: Leo <[EMAIL PROTECTED]> To: [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: [Fwd: Re: RSA Broken ??] Dear Barry Wels, For your information and further discussion and scrutiny. Here is a copy of my communication with Ron Rivest. The format of the illustration below seems to have been modified by my email software. Best regards, Leo -------- Original Message -------- Subject: Re: RSA Broken ?? Date: Sun, 04 Feb 2001 06:59:11 +0800 From: Leo <[EMAIL PROTECTED]> To: Ron Rivest <[EMAIL PROTECTED]> References: <[EMAIL PROTECTED]> Dear Ron, Thank you very much for the clarification and comprehensive explanation of the math concepts behind RSA. I really appreciate the time you spend on this communications. I just would like to get additional view and maybe some idea on comparative speed of the process I proposed of multiplying N by Y such that the product is all bit 1 to solve 2^X = 1 mod N. This, I believe, is really a "NEW" approach to find the private key. The "NEW" here is the Specific Use of 2 and the Use of simple Add-Shift-Compare binary operations to find X. To illustrate further: Say N = 55 = 110111 By repeated addition and shifting, we arrive at Y = 19065 = 100101001111001 110111 = bit 0 of Y = 1 + 110111 = bit 3,2,1 of Y = 100 -------------------- 111101111 + 110111 = bit 4 of Y = 1 --------------------- 10101011111 + 110111 = bit 5 of Y = 1 ----------------------- 110000111111 + 110111 = bit 6 of Y = 1 ------------------------- 1100111111111 + 110111 = bit 9,8,7 of Y = 100 ------------------------ 1000011111111111 + 110111 = bit 11, 10 of Y = 10 --------------------------- 100011111111111111 + 110111 = bit 14,13,12 of Y = 100 ------------------------------- 11111111111111111111 = 20 bits of 1, so X = 20 I believe this is a very fundamental function of a computer (add, shift and compare). Therefore, I believe this is a lot faster than a multiplication or division factoring approach. I just don't have any clue on how to compare this speed with "number sieves". Although it seems like a factoring approach for N, the operation is forward (addition) and one time (no trial and error). So, I believe should be very fast. My estimate is at least = LOG(N,2) * 2 faster than division factoring. To help me compare, maybe, you can give a clue on how fast is a "number sieve" approach compared with simple division. Again, thank you very much for your valuable time spent on this. I really appreciate it. Best regards, Leo Ron Rivest wrote: > Dear Leo -- > > Thanks for the more detailed explanation of your approach to attacking > RSA given in your emails (copied below). For the reasons I will > explain, and as you are perhaps aware, I think your approach is > unlikely to work in practice against large RSA numbers. It would be > very premature or misleading to characterize RSA as "broken" based on > your work to date. > > On the other hand, the strength of a cryptosystem can only be > determined by subjecting it to extensive analysis and attack by all > interested parties. I encourage you (and others) to try and find a > "shortcut" for breaking RSA faster than the known attacks (which also > don't work in practice when the numbers are large). Perhaps you will > someday find an attack that works efficiently. (Of course, I rather > hope you won't, but I don't want to discourage you from trying!) > > Let's go through some of the math and complexity issues. > > To begin with, you note that there may be several exponents that > work at decryption, and that the set of possible decryption exponents > may depend on the message. This is well-known. > > Let n = p*q, where p and q are prime. > > Let m be a "message", i.e. a number in the range 0 to n-1. > > For an arbitrary positive integer s and an integer r in the range > 0 to s-1, define order(r,s) as the number of distinct values that > occur in the sequence generated by r by multiplication: > r, r^2, r^3, ... > (all modulo s). Thus > order(2,5) = 4 as 2 generates the sequence 2,4,3,1,2,4,3,1,... (mod 5) > order(2,8) = 3 as 2 generates the sequence 2,4,1,2,4,1,... (mod 8) > order(3,6) = 1 as 3 generates the sequence 3,3,3,3,... (mod 6) > order(0,5) = 1 as 0 generates the sequence 0,0,0,... (mod 5) > One can also call order(r,s) the order of the subgroup generated by r, modulo s. > (I note for the record here that this definition is a slight extension of > the more usual one in that it covers the case that r is not relatively > prime to s.) > > When r and s are relatively prime, we have > r^order(r,s) = 1 (modulo s) > and order(r,s) is the least positive exponent for which this works. > > In any case we have that for any nonnegative integer k) > r^(1+k*order(r,s)) = r (modulo s) > and (1+k*order(r,s)) are the only exponents for which this works. > > Thus, for a message m modulo n, we have that > m^(1+k*order(m,n)) = m (mod n) > for all nonnegative integers k. > > For RSA encryption/decryption to work with message m, we need that > c^d = m (mod n) > where c is the ciphertext > c = m^e (mod n) > In other words, we need that > m^(ed) = m (mod n) > So it must be the case that > ed = 1 + k*order(m,n) > for some nonnegative integer k. > > What values can order(m,n) have? > > It is known that when p is prime > order(m,p) > must be a divisor of p-1, and that for any divisor v of p-1 > there is some message m such that order(m,p) = v. More > precisely, the number of messages m modulo p such that > order(m,p) = v is precisely equal to phi(v) when v>1, and > equal to 2 when v=1, where phi(v) is the Euler totient > function of v (the number of positive integers less than v that > are relatively prime to v). When v=1, the elements 0 and 1 > both have order(0,p)=order(1,p)=1. > > In the interesting case that p is a Sophie-Germain prime, so that > p = 2p'+1, where p' is also prime, then > p-1 = 2p' > and all elements have order 1, 2, p' or 2p'. Furthermore > 2 elements have order 1 (these are 0 and 1) > 1 element has order 2 (this is p-1, i.e. -1 mod p) > p'-1 elements have order p' > p'-1 elements have order 2p' > so all but the three elements 0, 1, and -1 (which have the obvious > orders 1, 1, and 2) have order divisible by p'. It is often > recommended that users of RSA use such primes. > > Now for RSA, when n = pq, we have > order(m,n) = lcm(order(m,p),order(m,q)) > where lcm(a,b) = ab/gcd(a,b) = the least common multiple of a and b. > > In the case that p and q are Sophie-Germain, p=2p'+1, q=2q'+1, where > p' and q' are prime, then the possible values of > order(m,n) > are > 1 > 2 > p' > 2p' > q' > 2q' > p'q' > 2p'q' > There are four elements of order 1 and eight elements of order 2. All > other orders have order divisible by p' or q', and so are large and very > hard to find without knowing p or q. > > To find a decryption exponent d that works for a message m, we can > pick and d satisfying > ed = 1 + k*order(m,n) > as noted above. This is equivalent to: > d = e^{-1} mod(order(m,n)) (*) > Given the possible variation in the values of order(m,n) and the possible > variation in the choice of k above, we see that there can be variation in > decryption exponents that work. > > However, if we want to choose a decryption exponent that works for *all* > messages m, then we need to use the fact that > order(m,n) is a divisor of lcm(p-1,q-1) > and that all such orders are possible. Denote > lambda(n) = lcm(p-1,q-1) > Then if we solve the equation > d = e^{-1} mod (lambda(n)) (**) > we get a result that satisfies (*) for all messages m. This is the usual > procedure. It is also the case that solving > d = e^{-1} mod (phi(n)) > where > phi(n) = (p-1)*(q-1) > will give a solution that also satisfies (**) and thus satisfies (*) > for all m. Working mod lambda(n) can save a little bit over working modulo > n, but typically not much, since gcd(p-1,q-1) is typically small. > > This explains your observation that there may be several decryption > exponents that work, and that even some decryption exponents work for > some messages and not for others. In the case of large Sophie-Germain > primes, however, almost all (all but approximately p+q) messages will > have order(m,n) divisible by p'q', and the set of decryption exponents > that "work" for m will satisfy > e*d = 1 (mod p'q'). > > Now, on to your proposed attack method, which is, as you note, not so > efficient for a large n. I would like to argue here that your method > is not likely to be efficient, given the state of the art. > > You propose first finding an x such that > 2^x = 1 (mod n) (***) > and then proceeding from there. > > ** This is a "known hard problem" with no good solutions known, even > ** though many have tried. It is known as the "discrete logarithm problem" > ** and appears to be just as hard as factoring the modulus n. > > Using the above notation, we must have that x is a multiple of order(2,n) > for (***) to hold. In the case that p and q are Sophie Germain, this > means that you are finding p'q' or a small multiple of p'q'. Suppose that > you actually found p'q' this way. Then of course > n = pq = (2p'+1)(2q'+1) = 4p'q' + 2p' + 2q' + 1 > so you could then compute > (n - 4p'q' - 1)/2 = p'+q' > Now, given the product p'q' of p' and q', and the sum p'+q' of p' and q', > it is easy to compute p' and q' separately. This means you have factored n, > since p and q are 2p'+1 and 2q'+1. > > Thus, in order to get started on your attack method, you need to solve > the discrete logarithm problem, which is believed to be hard, and is > likely to lead (as you see) to a factorization of n. Indeed, it is > known that finding any multiple of lambda(n) can lead to a > factorization of n, so that you really are doing nothing more than > proposing factoring n as a way of getting started on breaking RSA. > > The "usual division approach" for factoring n that you mention is of course > not the best approach to factoring n, and not what you should be considering > as a "competitor" to your approach. Indeed the "number field sieve" and its > generalizations can factor a number n in time > exp(c (ln n)^(1/3) (ln ln n)^(2/3) > where c = 1.92 and ln(x) is the natural logarithm of x. This is *much* faster > than trial division, but is not fast enough to allow factorization of > 1000-4000 bit numbers... > > While you thought you were attempting some "new" way of attacking RSA, > the method you propose is in fact equivalent to the "old" way of > attacking RSA: factoring the modulus n. There doesn't seem to be any > new leverage you are getting by thinking about the problem in the way > you propose. There is no reason to think that finding an x which is a > multiple of order(2,n) is any easier than directly finding the factors > p and q of n. (Indeed, order(2,n) must be divisible by p'q' in the case > that p and q are Sophie-Germain...) > > So, I hope this helps you think more clearly about the issues raised > in your proposed attack. > > There are many good references for this sort of thing; > one of my favorites is: > > book{Pomerance90, > editor = {Carl Pomerance}, > title = {Proc.\ of the {AMS} Symposia in Applied Mathematics: > Computational Number Theory and Cryptography}, > publisher = {American Mathematical Society}, > year = {1990} > } > > I think it is still in print. > > Some minor notes: I think properly when p=2p'+1 we should call p' > the Sophie-Germain prime and not p. And the sort of reasoning applied > above using Sophie-Germain primes is a good guide for you, but the > arguments against your approach would apply in any case, as long as > p-1 and q-1 have large prime factors (which they are likely to do in > general, even if p and q are chosen essentially at random). > > Cheers, > Ron Rivest > > ============================================================================== > To: Ron Rivest <[EMAIL PROTECTED]> > From: Leo de Velez <[EMAIL PROTECTED]> > Subject: Re: "RSA Broken" ?? > Reply-To: [EMAIL PROTECTED] > CC: [EMAIL PROTECTED] > Date: Sat, 3 Feb 2001 16:16:42 +0800 > X-Sender: [EMAIL PROTECTED] > X-Originating-Host: cache3.mydestiny.net [202.8.224.87]; Sat, 3 Feb 2001 08:19:10 GMT > X-Browser: Mozilla/4.7 [en] (WinNT; I), JavaScript: On > Content-Type: text/plain; charset=iso-8859-1 > > Dear Ron, > > I was surprised by the article because my agreement with Edu Lopez > over the phone was that I will email him the explanation > after i discuss the findings with you. Anyway, without the knowledge > of this press report, I sent you this morning the approach > that I am looking at. > > It was basically finding X such that 2^X = 1 mod N > > After the hard part above, > then, finding Y such that X * Y = 1 mod E (using extended euclidian > algorithm) > > and then, finding D = ( X * Y + 1 ) / E > > As I have said before, D is not necessarily equal to private key > (d). This means, other keys can decipher the code. The > decoding may not be 100% but based on my few tests, it is above 80%. > Another thing is X is a factor of (p-1)(q-1). It is at > least one half but usually one fourth or one sixth or one eight. > This means that the true private key can be calculated easily by > > multiplying X by two's and three's. > > I also stated in my email this morning that I am investigating two > approaches to find the X in 2^X = 1 mod N. > > One is using the Linear Feedback bit Shift Register. > The intention is to find a number Z such that Z * N = all binary > bits are equal to 1. This is done by repeated addition of binary > > code of N to N while moving the addend to the left such that the > bit 1 of addend is in line with the 0 bit of the partial sum. The > > number Z is the decimal representation of the shifting of the addend. > Then Z*N + 1 is a perfect log witch is the number of > resulting binary digits. This is now the X that I am looking for. > > As I have said, the number of operations involved is in the order > of 2^N / 5. Even though each operation is very fast because it > is binary and simple, the number of operations is enourmous. But > it will still be faster by about = 2 * LOG(N,2) versus a > simple division factorization. > > The other approach is to use the LOG(N,2) > > Starting from the Square(integer(square root of N)) and going down > by jumps of LOG(N,2). The work can be broken up to > different computers. > > The intention again is to find 2^X = 1 mod N but the testing of > X is in jumps of LOG(N,2). Once 2^X mod N is equal to a > discrete LOG of 2, (2^X = 2^y mod N), then we just subtract y from > X to find the usuable X that will give 2^X = 1 mod N. > This will increase the speed of factorization by at least LOG(N,2) > times versus normal division factorization approach. > > Then, by using the similar formula above, D can be calculated very > fast and again D is not necessarily equal to the private key > (d). As mentioned, it implies that other key can read the coded > message but not 100%. > > When I sent you my previous emails, I was still hoping that the extended > euclidian algorithm will allow me to approach a good > candidate for X which is a factor of (p-1)(q-1). However, as I use > bigger numbers, the number of iterations became larger. > The second approace was just a more stable replacement to that. > > I am sure there is a faster way to find 2^X = 1 mod N out there and > I would appreciate if you can help me on this. What I > believe I can claim as mine is the 2^X = 1 mod N approach to find > d and in so doing, finding also D that can replace d at least > 80% of the time. The serious implication is that the private key > is not unique at least for 80% of the messages. > > Best regards, > Leo de Velez > > At Friday, 2 February 2001, you wrote: > > >Hello again Leo -- > > > >I heard from a colleague that you have "gone public" with your claim > >to have a technique for decoding RSA ciphertexts (the information I > >received is copied below for your information). > > > >It is curious to compare the situation here with similar claims that I > >received from a mathematician in Japan. Fortunately for him, he made > >the details of his proposed method available to me before going public > >with his conjectured claim, and I was able to explain the error in his > >thinking to him before he "went public"... > > > >Perhaps you really do have a new technique here. If so, there are many > >people (including myself) who would be interested to see you provide > >some justification for your claims. Unfortunately, researchers in the > >field have seen all too many such claims evaporate when the details > >are made public. (Perhaps you recall, for example, the claims made by > >the Irish teenager Sarah Flannery a couple of years ago that received > >so much press attention.) So mere claims have little credibility > >until they are backed up by publication of the details and review by > >others in the field. > > > >So, let me encourage you to proceed down the path you have started, > >and make public or post the details of what you think you can do. (If > >you prefer, I can look them over for you before you publish, but of > >course this is your show and you might just want to publish or post > >what you have.) If you would like a "challenge example" to work on, I > >could generate one. (How large would you like it to be?) > > > >Your note to me earlier implied that you could thought you decrypt a > >significant fraction of the ciphertexts, but perhaps not all. Are you > >aware that the multiplicative properties of RSA then would imply that > >you could decrypt any ciphertext in polynomial expected time? (To > >decrypt ciphertext c, generate a random plaintext/ciphertext pair > >(p',c') using the public key, and decrypt c*c' with your technique, > >then divide the result by e' to obtain the desired ciphertext p > >corresponding to the target ciphertext c. Repeat as necessary.) > > > >Anyway, I look forward to hearing more. I encourage you to publish > >what you have and take your congratulations or lumps, as > >appropriate... Of course, please keep me informed of what you do, and > >feel free to contact me with details if you want feedback directly > >from me... And don't be too embarassed if the details don't actually > >work out in the end--you wouldn't be the first. The truth will win > >out in the end, for all of us... > > > > Cheers, > > Ron Rivest > > > >============================================================================== > >http://www.mb.com.ph/INFO/2001-02/IT020201.asp > > > >Friday, 2 February 2001 > >Pinoy math enthusiast finds fast way to decode RSA encryption > >By EDU H. LOPEZ > >A Filipino mathematics enthusiast has developed a new method of > decoding RSA > >(RivestShamir-Adleman) encryption using three simple formulas. > > > >Leo de Velez has discovered these three formulas are simple forward > >equations that allow fast decoding of RSA encryption. > > > >RSA is an Internet encryption and authentication system that uses an > >algorithm developed in 1977 by Ron Rivest, Adi Shamir, and Leonard > Adleman. > > > >The RSA algorithm is the most commonly used encryption and authentication > >algorithm and is included as part of the Web browser from Netscape and > >Microsoft. > > > >It's also part of Lotus Notes, Intuit's Quicken, and many other > products. > >The encryption system is owned by RSA Security. The company licenses > the > >algorithm technologies and also sells development kits. > > > >The technologies are part of existing or proposed Web, Internet, and > >computing standards. > > > >Here's how the RSA system works. The mathematical details of the > algorithm > >used in obtaining the public and private keys are available at the > RSA Web > >site. > > > >Briefly, the algorithm involves multiplying two large prime numbers > (a prime > >number is a number divisible only by that number and through additional > >operations deriving a set of two numbers that constitutes the public > key and > >another set that is the private key. > > > >Once the keys have been developed, the original prime numbers are > no longer > >important and can be discarded. > > > >Both the public and the private keys are needed for encryption/decryption > >but only the owner of a private key ever needs to know it. > > > >Using the RSA system, the private key never needs to be sent across the > >Internet. The private key is used to decrypt text that has been > encrypted > >with the public key. > > > >Thus, if I send you a message, I can find out your public key (but > not your > >private key) from a central administrator and encrypt a message > to you using > >your public key. > > > >When you receive it, you decrypt it with your private key. In addition > to > >encrypting messages (which ensures privacy), you can authenticate > yourself > >to me (so I know that it is really you who sent the message) by > using your > >private key to encrypt a digital certificate. When I receive it, > I can use > >your public key to decrypt it. > > > > Date: Sat, 03 Feb 2001 12:07:59 +0800 > From: "leo" <[EMAIL PROTECTED]> > To: "Ron Rivest" <[EMAIL PROTECTED]> > Subject: new way to decrypt RSA code > X-MDaemon-Deliver-To: [EMAIL PROTECTED] > X-Return-Path: [EMAIL PROTECTED] > > Dear Ron, > > The solution to 2^D = 1 mod N > > There are two approaches that I am looking at. > > The first is to multiply the binary code of N with another number so > that the result will all be binary 1. This means adding the binary code > of N while shifting to the left. This will give a number and add 1 will > give a power of 2 (log base 2 is discrete). The discrete log is a > factor of (p-1)(q-1). At least one half, usually one forth, one sixth, > or one eighth of (p1-)(q-1). > > For N=55 = 110111 > Multiple = 19065 = 100101001111001 > > Although the operation is simple and fast (add and shift), the number of > operation is in the order of 2^N / 5 . The operation is also not easy > to distribute. > > Another approach is to use the log N base 2. > > Using Nroot = [sqr(int(sqrtN))] and Ndelta = [N-sqr(int(sqrtN))] > and L = int(LOG(N,2)) > > i=1 > Starting point X = Nroot - i * Ndelta > j = 0 to Ndelta/LOG(N,2) > Y = X + j * L > Calculate Z = LOG((MOD((2^Y),N),2) > If Z is discrete, the D = Y - Z (check 2^D = 1 mod N) > > i=2 (another computer doing this) > Starting point X = Nroot - i * Ndelta > j = 0 to Ndelta/LOG(N,2) > Y = X + j * L > Calculate Z = LOG((MOD((2^Y),N),2) > If Z is discrete, the D = Y - Z (check 2^D = 1 mod N) > > i= up to int(N/Ndelta) (another computer doing this) > Starting point X = Nroot - i * Ndelta > j = 0 to Ndelta/LOG(N,2) > Y = X + j * L > Calculate Z = LOG((MOD((2^Y),N),2) > If Z is discrete, the D = Y - Z (check 2^D = 1 mod N) > > I would appreciate if you can share with me a faster way of finding > > 2^D = 1 mod N > > Best regards, > Leo > > -----Original Message---- > From: Ron Rivest <[EMAIL PROTECTED]> > Date: Wed, 31 Jan 2001 00:49:09 -0500 > Subject: [[EMAIL PROTECTED]: Re: [[EMAIL PROTECTED]: new way to decrypt > RSA code]] > > > Support TEAMMAIL. Be a Panelist in our Surveys. Email to > > [EMAIL PROTECTED] > > > > > > I don't understand your explanation at all. What is y? What is x? > > > > Maybe you could explain how it works on the following example? > > n = 55 > > e = 3 > > ciphertext = 2 > > > > Thanks... > > > > Thanks, > > Ron Rivest > > > > ------- Start of forwarded message ------- > > Date: Mon, 22 Jan 2001 13:44:03 +0800 > > From: "leo" <[EMAIL PROTECTED]> > > To: "Ron Rivest" <[EMAIL PROTECTED]> > > Subject: Re: [[EMAIL PROTECTED]: new way to decrypt RSA code] > > X-MDaemon-Deliver-To: [EMAIL PROTECTED] > > X-Return-Path: [EMAIL PROTECTED] > > > > Dear Ron Rivest, > > > > Thank you for your reply. > > > > The first and second equations are similar to the equation used to > > calculate the secret key d from e and (p-1)(q-1). One difference is > > they do not need (p-1)(q-1). > > > > And the third equation is a simple z= y/x, where z is the secret > > code. > > BUT z is not necessarily equal to d. This means secret codes are not > > unique. > > > > I'm only using a pentium 100 laptop and an excel software to decipher > > secret key of N (value only up to 10^8). Its fast and instant. > > > > Best regards, > > Leo de Velez > > > > > > > > > > - -----Original Message----- > > From: Ron Rivest <[EMAIL PROTECTED]> > > Date: Sun, 21 Jan 2001 21:39:30 -0500 > > Subject: [[EMAIL PROTECTED]: new way to decrypt RSA code] > > > > > Support TEAMMAIL. Be a Panelist in our Surveys. Email to > > > [EMAIL PROTECTED] > > > > > > > > > I'd be curious to hear more---how does your method work?? > > > > > > Ron Rivest > > > > > > > > > ------- Start of forwarded message ------- > > > From: "teammail" <[EMAIL PROTECTED]> > > > To: <[EMAIL PROTECTED]> > > > Subject: new way to decrypt RSA code > > > Date: Sat, 20 Jan 2001 17:02:41 +0800 > > > Content-Type: multipart/alternative; > > > boundary="----=_NextPart_000_0007_01C08302.CCE16270" > > > X-Priority: 3 > > > X-MSMail-Priority: Normal > > > X-MimeOLE: Produced By Microsoft MimeOLE V5.50.4133.2400 > > > X-MDaemon-Deliver-To: [EMAIL PROTECTED] > > > X-Return-Path: [EMAIL PROTECTED] > > > > > > This is a multi-part message in MIME format. > > > > > > - ------=_NextPart_000_0007_01C08302.CCE16270 > > > Content-Type: text/plain; > > > charset="iso-8859-1" > > > Content-Transfer-Encoding: quoted-printable > > > > > > Im Leo de Velez residing at 26B Prudent Lane, Sanville Subd, Quezon > > = > > > City, Philippines=20 > > > with phone number 63-917-532-9297. Im a math enthusiast and i > > found > > > a = > > > new way to=20 > > > decrypt RSA encryption. The process involved three simple > > formulas. > > > I = > > > have tried it=20 > > > on small values of N and it worked well and fast. It basically > > > exploits = > > > the=20 > > > non-primeness of (p-1)(q-1). Using one of its factors, the secret > > > key = > > > can be calculated=20 > > > (forward direction).=20 > > > > > > - ------=_NextPart_000_0007_01C08302.CCE16270 > > > Content-Type: text/html; > > > charset="iso-8859-1" > > > Content-Transfer-Encoding: quoted-printable > > > > > > <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> > > > <HTML><HEAD> > > > <META http-equiv=3DContent-Type content=3D"text/html; = > > > charset=3Diso-8859-1"> > > > <META content=3D"MSHTML 5.50.4134.600" name=3DGENERATOR> > > > <STYLE></STYLE> > > > </HEAD> > > > <BODY bgColor=3D#ffffff> > > > <DIV><FONT face=3DArial size=3D2> > > > <DIV><FONT face=3DArial size=3D2>Im Leo de Velez residing at 26B > > > Prudent = > > > Lane,=20 > > > Sanville Subd, Quezon City, Philippines <BR>with phone number=20 > > > 63-917-532-9297. Im a math enthusiast and i found a new way to > > > = > > > <BR>decrypt=20 > > > RSA encryption. The process involved three simple > > > formulas. = > > > I have=20 > > > tried it <BR>on small values of N and it worked well and fast. > > > It=20 > > > basically exploits the <BR>non-primeness of (p-1)(q-1). Using > > > one = > > > of its=20 > > > factors, the secret key can be calculated <BR>(forward > > direction).=20 > > > </FONT></DIV></FONT></DIV></BODY></HTML> > > > > > > - ------=_NextPart_000_0007_01C08302.CCE16270-- > > > ------- End of forwarded message ------- > > ------- End of forwarded message

- Pinoy math enthusiast finds fast way to decode RSA encrypti... Marc Branchaud
- Re: Pinoy math enthusiast finds fast way to decode RSA... Bill Stewart
- Barry