RSA encryption and decryption in Python using OpenSSL

Question

I have public key, private key and encrypted message ( e, n, d, message). I want to encrypt and decrypt with OpenSSL in Python.

I have keys I don't need to generate them. I saw many questions about RSA but all of them creates keys, by generate method. I haven't seen questions about performing encryption and decryption.

How do I perform RSA encryption and decryption in Python usomg OpenSSL?

What I have is the followings:

#Public key:
e="65537"

n="2483790199491205341506001624338547531741771200963322451" \
  "16318655098567854682220895878748602513720919196015584783557" \
  "704400541915850094004767776687012394493261448676783015273985771" \
  "1238030112386627501698642461625407366533907839206541565912321199" \
  "009791795944233570230631191423356738502486763195167267521973031" \
  "6838578210434343067511636079081818744400113533624136339709745782" \
  "32161853372590090308494113224155565481298018056338822080805188480" \
  "139150684063550507331062187412707210886548924698896783031493679037" \
  "3122088161029787856707927049345768779125257912445784686277424030038" \
  "539380288863347855630618237433032833865316901740219"


#Private key
d="152501997096795110757068525202189319208625862269501399381045003" \
  "01373684948285528219578200125958795897780598922907027278290745917" \
  "4083840545807194541888429655727807270271016523695687179904011971106" \
  "46024638603131783118232131092639581621182826911051011196270811088775" \
  "8622627957416117004996969971673524599345136221501081814180958260506967" \
  "05549363779862358358393233189560520163106785535319492545898745183439" \
  "10980478364023104227720426942196244946117979269924656213962726626606" \
  "77452212629548965644705371048342816305068001182195025882564173365857" \
  "07762540909960941277936950557159506459454566798472128560135656506235" \
  "741389170953"


#encrypted message
encoded="187216163520278606105320112446137004408231369834741341053563682" \
        "277774349916058822189964158715390402738262899525931062389534962" \
        "09104749822344117450601254708536373034264130933521987327974000" \
        "255146756518397668069770185737907343422454676477169144712992560" \
        "738066894543224559303296179944700852861503983647039123452966586" \
        "43024446530008588087574157621730825724439869400851215840977916" \
        "767440706251849931986529460039147463908090086303953826751056882" \
        "5732583473943114017472152320746478960753673137088195122814398113" \
        "5288648561417818449968250721180493107501204327582989947582671" \
        "70231934908068721013345590521202959891172540575563129"

You need to show what format your key and message is in and what you've tried. As it stands the question is off topic (for at least 3 different reasons). Besides that, if shown a working option, how would you know it is secure? — Maarten Bodewes
– Maarten Bodewes, Commented Oct 29, 2016 at 19:26
Well, you seem to be right. I'm getting fodder when searching for Python and RSA_encrypt and RSA_decrypt. Maybe PyOpenSSL lacks the bindings for the functions. I thought this topic would have a few good posts and answers. — jww
– jww, Commented Oct 29, 2016 at 19:53
@jww yeah, actually there is a method called importKey which imports the key from the file and I put my key in the file, but my attempts were failed — eneski
– eneski, Commented Oct 29, 2016 at 20:00
@MaartenBodewes Thanks for warning, I put what I got (keys and cipher text) but I tried so many things and I can not put them here, it would make the question unreadable, though — eneski
– eneski, Commented Oct 29, 2016 at 20:02
@MaartenBodewes With three arguments, pow() is a modular operation. — r3mainer
– r3mainer, Commented Oct 30, 2016 at 0:05

r3mainer · Accepted Answer · 2016-10-30 13:35:28Z

1

The ciphertext can be decoded without using pyOpenSSL or any other external libraries.

Simply reverse the encryption operation using the formula m = c^d mod n as follows:

n="24837..."
d="15250..."
encoded="18721..."
plaintext = pow(int(encoded,10),int(d,10),int(n,10))
print hex(plaintext)[2:-1].decode('hex')

The pow() operator performs a modular exponentiation operation to calculate (encoded**d) % n. In the print statement, the resulting value is converted to hexadecimal (removing the first two characters 0x and the last character L), and then decoded as a hexadecimal string to retrieve the original bytes.

Since the message was encoded without using any padding, no further operations are required.

answered Oct 30, 2016 at 13:35

r3mainer

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