RSA Decryption given n, e, and phi(n)

Question

my cryptography professor gave us this problem for extra credit a while back and I attempted it but I didn't get it correct. I have gone back to it, but I'm even more lost now than I was the first time (HEADS UP: my professor gave us some long numbers to work with)

Let N=

 217480967426598493570186980401996167920452820950992854687035132
 726376073118953668642571927853352339590583090604658684239518232
 853572979131254064492477407811878270605929141044977950456991658
 882119063692415321357447387047198644556238539408419478596527501
 23329821235383771649185149402914522002819011319590369529


 And e=65537 be a public key for the RSA cryptosystem

The ciphertext is:

503502864628940396744635609090061402472498491194463969923479322
089849770618612322621019979098961384995354204702163333139805707
759660081519394083069273037638282947420860920004667666344095765
710257079484209928467972889843783133155796096794854080979925590
2703014867201045003016001267189341232653910252303505881

And we determine that:

 phi(n)=
 217480967426598493570186980401996167920452820950992854687035132
 726376073118953668642571927853352339590583090604658684239518232
 853572979131254064492477381836798664822705327075966772445783246
 838640744314191496615629393761046078697812048683249333166663641
 55001553295827687037711471146642763970634674131635418276

I was supposed to decode the message (The message was converted into a number by writing it in base 256 using the ASCII codes for the individual characters)

If anyone could provide a step by step explanation of how I go about doing this, I would greatly appreciate it.

Use a big integer library, most programming languages either contain one (at least C# and Python do) or have a readily available third party implementation. (unfortunately the wolfram alpha link above doesn't work, it truncated phi) — CodesInChaos, Oct 30 '14 at 20:19
@CodesInChaos: Once I've done the calculations, how do I use d and m to decrypt? — Jok3r, Oct 30 '14 at 20:22
Well, do you understand how RSA encryption and decryption work? — pg1989, Oct 31 '14 at 01:14
I warmly suggest to use ipython or, if you're a mathematician, Pari/GP or Sage. You can do most of cryptography computation very easily, as it was you pocket calculator. — ddddavidee, Oct 31 '14 at 06:42
If this is more about programming side, you need to ask more explicitly about manipulating large numbers in Stack Overflow, not here. — Ray, Nov 01 '14 at 18:15

score 3 · Answer 1 · edited Oct 31 '14 at 15:08

3

We have $d = e^{-1} \pmod{ \phi(N) }$, this implies that $m = c ^ d \pmod{N}$

edited Oct 31 '14 at 15:08

poncho

147,019
11
229
360

answered Oct 31 '14 at 02:45

higgs

31
1

sadnoe · Accepted Answer · 2014-10-31T23:08:03.493

I'm sorry @J0ker, I found new answer (maybe there are errors value in my first answer).

My new answer $m =$

49314552466695255586203088029816774295110376055124609941914798033775741215800363731230533018093001338140450279336308798327354131807371119497156895131357788895448541113895626439002123851

I try to convert $m$ to base 256

[75, 110, 111, 119, 108, 101, 100, 103, 101, 32, 105, 115, 32, 112, 111, 119, 101, 114, 46, 32, 80, 108, 101, 97, 115, 101, 32, 117, 115, 101, 32, 116, 104, 101, 32, 112, 111, 119, 101, 114, 32, 121, 111, 117, 32, 104, 97, 118, 101, 32, 102, 111, 114, 32, 103, 111, 111, 100, 44, 32, 97, 110, 100, 32, 110, 111, 116, 32, 102, 111, 114, 32, 101, 118, 105, 108, 46]

with ASCII code

Knowledge is power. Please use the power you have for good, and not for evil.

try this (in Java Language)

import java.math.BigInteger; import java.util.ArrayList;

public class RSADecrypt {

public static void main (String[] args) {

    BigInteger N,phiN,e,d,m,c;

    // chipertext c, plaintext m

    N = new BigInteger ("21748096742659849357018698040199616792045282095099285468703513272637607311895366864257192785335233959058309060465868423951823285357297913125406449247740781187827060592914104497795045699165888211906369241532135744738704719864455623853940841947859652750123329821235383771649185149402914522002819011319590369529");

    e = new BigInteger ("65537");

    c = new BigInteger ("5035028646289403967446356090900614024724984911944639699234793220898497706186123226210199790989613849953542047021633331398057077596600815193940830692730376382829474208609200046676663440957657102570794842099284679728898437831331557960967948540809799255902703014867201045003016001267189341232653910252303505881");

    phiN = new BigInteger ("21748096742659849357018698040199616792045282095099285468703513272637607311895366864257192785335233959058309060465868423951823285357297913125406449247738183679866482270532707596677244578324683864074431419149661562939376104607869781204868324933316666364155001553295827687037711471146642763970634674131635418276");

    d = e.modInverse(phiN);
    m = c.modPow(d, N);

    System.out.println("d = "+d);           
    System.out.println("m = "+m);

    System.out.println("m in base 256 = "+base256(m));
    System.out.println("Convert with ASCII \n"+ Encode256(base256(m)));

}

static ArrayList<BigInteger> base256 (BigInteger M) {
BigInteger base = new BigInteger("256");
    ArrayList<BigInteger> message256 = new ArrayList<BigInteger>();
BigInteger sisa=M;
BigInteger k;
double z = Double.parseDouble(M.toString());
double p = Math.floor(Math.log(z)/Math.log(256));
int r = (int) p;
    for (int j=0;j<=r;j++){
        k=sisa.mod(base);
        sisa=sisa.divide(base);
        message256.add(k);
}
return message256;
}

static String Encode256 (ArrayList<BigInteger> ascii) {
String ascii256="";
int g,dmp;
for (int i=0;i<ascii.size();i++) {
g = Integer.parseInt(""+ascii.get(i));
ascii256=ascii256+( (char) g );
}
return ascii256;
}

}

this is my result from this code enter image description here

RSA Decryption given n, e, and phi(n)

2 Answers2

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