# Given n and e find p and q

In the following blogpost I will explain why it is a bad idea to use small RSA keys. To make things look and feel real, I will demonstrate all steps needed to factorize and recover a private key. Two keys are required to succesfully encrypt and decrypt a message.

A keypair consists of the following keys:.

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Let's have a short look on how the RSA key generation works:. The genrsa command does all the steps and maybe a bit more. Note that the modulus n is bit long. However, the private key is our secret and we need the public key to encrypt a message. Extract the public key with the -pubout switch:. As you can see, our public key contains only the modulus n and the exponent e. The file my. Note: This only works for messages which are smaller than the modulus.

Usually the message is encrypted with a symmetric key which is in turn encrypted with RSA. As you can see, we encrypted our message "Hi" and the result is gibberish. Only the recipient can decrypt it using his private key. As said above, the sender needs the recipient's public key to encrypt a message. Thus an adversary can recover the private key and decrypt the message. I have chosen a pretty small key of bits above. Let's assume we are the adversary and are interested in recovering the contents of the message.

Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. First, factor n. This is not hard; since sqrt is That will give you p and q.

You do not need C code for that; I did it by hand. The algorithm gives a negative result. Learn more. Asked 6 years, 11 months ago. Active 4 months ago. Viewed 13k times. Last Unicorn Last Unicorn 71 1 1 gold badge 1 1 silver badge 3 3 bronze badges.

It is believed to be difficult to obtain p and q from n, and there is no publicly known way to do it in a feasible amount of computer time for large n in general. The security of the RSA algorithm depends upon this. Active Oldest Votes.

It only takes a minute to sign up. From the definition of the totient function, we have the relation:. Using the quadratic equation shown above, we need to use the following coefficients for the equation:. In fact that's just what happens during normal RSA key generation. Wikipedia about RSA key generation :. We will treat them as our unknowns:. Let's start:. The security of this system needs to be examined in more detail. In particular, the difficulty of factoring large numbers should be examined very closely.

Once the method has withstood all attacks for a sufficient length of time it may be used with a reasonable amount of confidence. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Why is it important that phi n is kept a secret, in RSA? Ask Question.

Asked 7 years, 4 months ago. Active 5 months ago.

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Very well done. Peter Kluge Peter Kluge 51 1 1 bronze badge. Nik Bougalis Nik Bougalis 1 1 silver badge 6 6 bronze badges.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.

It only takes a minute to sign up. You are almost finished. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 7 years, 3 months ago.

Active 7 years, 3 months ago. Viewed 7k times. Anyone cares to explain. Patrick Li 3, 4 4 gold badges 17 17 silver badges 34 34 bronze badges. Favolas Favolas 7 7 silver badges 12 12 bronze badges. Active Oldest Votes. As you can see, you were almost there.

## How I recovered your private key or why small keys are bad  It only takes a minute to sign up. From there, your public key is [n, e] and your private key is [d, p, q]. Once you know those, you have the keys and can decrypt any messages - no cracking necessary! Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 7 years, 4 months ago.

Active 7 years, 4 months ago. Viewed 12k times. How would one find the secret key in a simple RSA encryption when given p, q and e? Active Oldest Votes. You've already been given everything you need to decrypt any messages. More details are available here. Polynomial Polynomial k 37 37 gold badges silver badges bronze badges. This kinda makes sense, though I had trouble with calculating d.

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Ended up using the extended Euclidean algorithm. KeithS I'm aware they don't call me Polynomial for nothing! The reason I mentioned the alternative form is that most implementations of large number libraries have the ability to raise an integer to an arbitrary exponent, but not as many have explicit division capabilities.

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Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It only takes a minute to sign up. I am given the qpand e values for an RSA key, along with an encrypted message. I tried calculating d with the Extended Euclidean algorithm, but came out as 1. How should I calculate d? I used the following python code to compute the private exponent and perform decryption. It uses the extended euclidean algorithm:. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. RSA given q, p and e? Asked 5 years, 6 months ago. Active 6 months ago. Viewed 28k times. Bob Bob 37 1 1 gold badge 1 1 silver badge 2 2 bronze badges. Scientific notation will not cut it, you need all the digits in order for it to work. Having said that, your decryption exponent is still not right. Code requests are off topic on Crypto. Active Oldest Votes. As far as my tests went this version handles all valid inputs properly. You need to explain this in your answer, provide a link to the previous answer or even comment, right click on the time behind the comment to obtain the link and of course indicate how your code solves the issue.

This could be a very helpful contribution if well applied. The Overflow Blog. Featured on Meta. Feedback on Q2 Community Roadmap. Do we need the theory tag. Related 9. Hot Network Questions. Cryptography Stack Exchange works best with JavaScript enabled.The binomial distribution is the basis for the popular binomial test of statistical significance. The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N.

If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distributionnot a binomial one. However, for N much larger than nthe binomial distribution remains a good approximation, and is widely used.

The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function :. The formula can be understood as follows. This k value can be found by calculating.

There is always an integer M that satisfies . M is the most probable outcome that is, the most likely, although this can still be unlikely overall of the Bernoulli trials and is called the mode.

The cumulative distribution function can be expressed as:. It can also be represented in terms of the regularized incomplete beta functionas follows: . Some closed-form bounds for the cumulative distribution function are given below.

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Suppose a biased coin comes up heads with probability 0. What is the probability of achieving 0, 1, This follows from the linearity of the expected value along with fact that X is the sum of n identical Bernoulli random variables, each with expected value p. This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:.

We find. In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:.

### The Binomial Distribution

Hoeffding's inequality yields the bound. However, the bounds do not work well for extreme values of p. In this case a better bound is given by .

Asymptotically, this bound is reasonably tight; see  for details.

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An equivalent formulation of the bound is. Both these bounds are derived directly from the Chernoff bound. It can also be shown that. This is proved using the method of types see for example chapter 11 of Elements of Information Theory by Cover and Thomas .

This estimator is found using maximum likelihood estimator and also the method of moments. It is also consistent both in probability and in MSE. A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. The Bayes estimator is biased how much depends on the priorsadmissible and consistent in probability.

This method is called the rule of successionwhich was introduced in the 18th century by Pierre-Simon Laplace.

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When estimating p with very rare events and a small n e. In such cases there are various alternative estimators. Even for quite large values of nthe actual distribution of the mean is significantly nonnormal. The notation in the formula below differs from the previous formulas in two respects: .

The exact Clopper—Pearson method is the most conservative.

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The Wald method, although commonly recommended in textbooks, is the most biased.