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 January 3rd, 2018, 08:10 PM #1 Newbie   Joined: Jan 2018 From: Cyberaid Posts: 2 Thanks: 0 Math Focus: geomerty and number theory Mathematical Attack on RSA Hello, I am trying to find an equation that will find p and q in the N = p * q problem of RSA cryptography. What I have posted here is my latest attempts. I am having trouble finding the error when a square root is calculated by a computer. Yes, I know this is an impossible problem. But hear me out. I have a long polynomial equation that seems more complicated than basic factoring by division. But as I show here the equation is more processing intensive, but tells whether a given x value is higher or lower, because the calculated PNP can be subtracted from the given PNP. So, we know N and plug in a p to find how far the plugged-in p creates an N that is a calculated distance from the given N. It sounds much more complicated than it actually is, but I just wanted to give the background of this problem I created. Please post your comments and/or questions. Code: PNP = 85 x = 85 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 85 161 Sqrt[4123/2] 1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4)) N[1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4))] -0.249277 PNP = 85 x = 5 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 5 Sqrt[4179323/2]/17 1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17]) N[1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17])] 1.37825 PNP = 85 x = 7 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 7 (11 Sqrt[45773587/2])/595 1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595]) N[1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595])] 1.88414 PNP = 85 x = 3 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 3 Sqrt[847772947/2]/255 3 + Sqrt[88 - Sqrt[847772947/2]/255] N[3 + Sqrt[88 - Sqrt[847772947/2]/255]] 5.69458 PNP = 85 x = 1 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 1 (7 Sqrt[13123/2])/85 1 + Sqrt[86 - (7 Sqrt[13123/2])/85] N[1 + Sqrt[86 - (7 Sqrt[13123/2])/85]] 9.90669
 January 9th, 2018, 05:32 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 I have no idea what you are trying to say here. In particular, if you are trying to find primes p and q such that pq= N, for given N, where does "PNP= 85" and "x= 85" come from? What is special about "85"? Are you saying that you are specifically setting N= 85 and trying to find p and q such that pq= 85? That's kind of trivial isn't it? 85= 5(17). How would this work for other "N"?
 January 9th, 2018, 08:05 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 Math Focus: Mainly analysis and algebra The whole point about RSA is that factorising a large semiprime is mathematically difficult. This means that it is possible but the best known algorithms take too long to solve the problem to be practicable. I'm not sure what speed the best algorithms are, but I'd guess at polynomial time or $n\log n$ which means that as soon as it becomes possible to factorise an $N$, we just need to pick bigger semiprimes to put factorisation safely out of reach. What I'm trying to say is that your first port of call should be to study modern factorisation techniques and research to understand the current state of knowledge on the subject.
January 9th, 2018, 10:26 AM   #4
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Quote:
 Originally Posted by Trurl Hello, I am trying to find an equation that will find p and q in the N = p * q problem of RSA cryptography. What I have posted here is my latest attempts. I am having trouble finding the error when a square root is calculated by a computer. Yes, I know this is an impossible problem. But hear me out. I have a long polynomial equation that seems more complicated than basic factoring by division. But as I show here the equation is more processing intensive, but tells whether a given x value is higher or lower, because the calculated PNP can be subtracted from the given PNP. So, we know N and plug in a p to find how far the plugged-in p creates an N that is a calculated distance from the given N. It sounds much more complicated than it actually is, but I just wanted to give the background of this problem I created. Please post your comments and/or questions.
What is PNP? No clue in what you have written. You did define a p and an N. Do they have anything to do with PNP?

What is x? No clue in what you have written. It appears to be an iterated guess.

Where does F come from? No clue in what you have written.

Lastly, you do realize that iterative methods for factoring numbers have been known for literally millennia, don't you? Finding a new one is significant only if you can show that, for some material class of cases, the new method is significantly faster than existing methods.

January 9th, 2018, 10:32 AM   #5
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Quote:
 Originally Posted by Trurl If [PNP - F > 0, x = x + Sqrt[PNP - F + x]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ]
If [PNP - F = 0 ??........]

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