Euler’s totient function is defined as the number of positive integers relatively prime to n (including 1). E.g. φ(12) = 4 ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), and φ(15) = 8. http://www.thescienceforum.com/mathematics/14111-modular-multiplicative-inverse-context-rsa.html https://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Extended_Euclidean_algorithm https://docs.google.com/viewer?url=www.math.utah.edu/~fguevara/ACCESS2013/Euclid.pdf as to why the extended Euclid’s algo can be used to find the modular multiplicative inverse https://docs.google.com/viewer?url=ftp://ftp.rsasecurity.com/pub/pkcs/pkcs-1/pkcs-1v2-1.pdf https://docs.google.com/viewer?url=ftp://ftp.rsasecurity.com/pub/rsalabs/rsa_algorithm/rsa-oaep_spec.pdf In particular, ^ page 9. OAEP is the padding scheme ( http://crypto.stackexchange.com/questions/10145/rsa-pcks1-v2-1-rsaes-oaep-algorithm http://crypto.stackexchange.com/questions/2074/rsa-oaep-input-parameters ) , whereas I2OSP and OS2IP (on page 4); what really helped things come full circle for me is realizing how they represent arbitrary data as an integer (first converting it to an octet string). Without further do, let’s test our generated keys by encrypting and decrypting the number 521 (any number smaller than 527, our modulus, will do. ( http://stackoverflow.com/questions/10061626/message-length-restriction-in-rsa ) )

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$ python
Python 2.7.5+ (default, Feb 27 2014, 19:37:08)
[GCC 4.8.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> print 521**41 % 527
346
>>> print 346**281 % 527
521

Note that if we try with a larger number…

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>>> print 1000**41 % 527
411
>>> print 411**281 % 527
473

Nope. What really helped things come full circle once again (or full sphere..) http://en.wikipedia.org/wiki/Pretty_Good_Privacy#Confidentiality http://superuser.com/questions/383732/how-does-ssh-encryption-work It makes more sense to use a symmetric encryption algorithm with high throughput to encrypt the data first, then use PKI to encrypt and transfer the key. And that is how the world works.

https://wido.me/sunteya/setup-a-socks-proxy-server-pass-by-secure-firewall/ http://www.bock.nu/blog/secure-firewall-bypass-danted-stunnel SSH Tunneling is TCP over TCP over a single connection. Rather unreliable. Socks can handle multiple connections in a non-blocking fashion. Also, to give you security on par with SSH, make sure to use AES256 somewhere in your cipher chain;

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transfusion@shell:~$ openssl ciphers -v -tls1 | grep 'AES(256)'
ECDHE-RSA-AES256-SHA384 TLSv1.2 Kx=ECDH     Au=RSA  Enc=AES(256)  Mac=SHA384
ECDHE-ECDSA-AES256-SHA384 TLSv1.2 Kx=ECDH     Au=ECDSA Enc=AES(256)  Mac=SHA384
ECDHE-RSA-AES256-SHA    SSLv3 Kx=ECDH     Au=RSA  Enc=AES(256)  Mac=SHA1
ECDHE-ECDSA-AES256-SHA  SSLv3 Kx=ECDH     Au=ECDSA Enc=AES(256)  Mac=SHA1
SRP-DSS-AES-256-CBC-SHA SSLv3 Kx=SRP      Au=DSS  Enc=AES(256)  Mac=SHA1
SRP-RSA-AES-256-CBC-SHA SSLv3 Kx=SRP      Au=RSA  Enc=AES(256)  Mac=SHA1
DHE-RSA-AES256-SHA256   TLSv1.2 Kx=DH       Au=RSA  Enc=AES(256)  Mac=SHA256

Alternatively, if you’re not sure what to use ,

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options = NO_SSLv2
ciphers = HIGH:MEDIUM

in your stunnel.conf should suffice.