Decoding RSA Problem: How to Crack Public Exponent and Modulo N

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The discussion focuses on the process of decoding an RSA-encrypted message using the public exponent and modulo N. The user initially struggles to compute the original message from the ciphertext and public key parameters, despite successfully calculating the private key. After several attempts and breaking down the calculations into manageable parts, the user discovers a critical error in their exponentiation. Correcting the mistake leads to successfully retrieving the original message, which translates to a Dutch phrase indicating understanding of the trick involved. The thread highlights the importance of careful verification in cryptographic computations.
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I just read up a bit on RSA and tried to do a problem I have in my algebra book. The goal is to crack/decode the message c=m^e mod N, obtained with public exponent e and modulo N.
I just puzzled away and tried a few things. I`m not a number theorist so my method may not be the most efficient, but it should work. However, my final answer doesn't match. When I compute m^e mod N it's not the same as I started with. Hopefully someone can spot the error.

Given is:

N = 15241578753238836751577503665157706318489955952973821

e = 31

c = m^e = 861510404336695227251078663863347293625350352438927

I just let Maple grind for a few minutes and it found the prime factorisation of N = pq:
p=1234567890123456789012345678901231

q=12345678901234567891
From which I find the totient:

\phi(N)=(p-1)(q-1)=15241578753238836750342935775034237183798709039504700

Then I found 'd' such that ed \equiv 1 \bmod \phi(N):

d = 11799931937991357484136466406478119110037710224132671

So now that I've found the private key d, I should just compute c^d and I should get m. However, Maple finds
c^d=861510404336695227251078663863347293625350352438927^{11799931937991357484136466406478119110037710224132671} \bmod N a bit hard to swallow (well, me too). So I thought about breaking d up into smaller pieces that Maple can handle and compute the parts manually. It goes fine as long as the exponent is not much larger than in the order of 50000.

I did: d=(11)(17)^2(20116367)(99732718879652629)(1850127955377590503746343)
Then raised c to the power 11, than to the power 17^2, all mod N ofcourse.
What I was left with is then:

m=10629575215834074693129969195052025048670006092854912^{(1850127955377590503746343)(99732718879652629)(20116367)}=x^{abc}
with

x=10629575215834074693129969195052025048670006092854912

a = 20116367 = 1+(2)(1193)(8431)

b = 99732718879652629 = 13+(8)(3)(59)(73)(21863)(2143)(20593)

c = 1850127955377590503746343 = 31+(8)(3)(1223)(3313)(19025787016433837)

So I've broken up a,b and c into smaller pieces again and repeated the scheme. The answer is below, I've introduced some symbols for clarity. Everything is assumed to be taken modulo N.

y = x^a = 1610044581396706062621617218007597816610136362872871

z = y^b = 8295213403069483304303211306255358878287773906357859

m = z^c = (z^13)(2882470640844826906680249819845717295058993606788553)^(19025787016433837)

For the last step I hacked up 19025787016433837 into 10+(49)(349)(25447)(21937)(1993) and putting it all together gave me:
m = 1632801515822720824329177013717453486198256157145735

But when I calculate m^e (mod N) I do not get the original c. So something is wrong, but I've doublechecked the whole thing a few times and I can't find a mistake. It may be something elementary, but I'd appreciate any help.

Ciao
 
Last edited:
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Doh!

I did z^13 instead of z^31. Silly me. After a few hours of checking I didn't notice that. And if I didn't make that Latex error I probably still missed it.

So with z^31 it does give the right answer:
m = 14210805021005040520182103202305120205071805160514

If you group the digits in pairs and let A=01, B=02, etc, it says
NUHEBJEDETRUCTWELBEGREPEN, which is (apart from spacebars and a spelling error) dutch for 'You've understood the trick by now'.
I hoped it would contain some ancient wisdom, but oh well.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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