Lua and RSA encryption - encryption

Question

I've been working on an RSA encryption script in Lua, with the assistance of BigNumbers (http://oss.digirati.com.br/luabignum/bn/index.htm), and I pretty much have a working code. I'm stuck, however, because in a small percentage of cases, the encrypted original message is not decrypted to the original message, and I cannot figure out why. Please note that this will deal with very large numbers (1.08e107, for example). The entire code I've written is below, but here's a breakdown of what it should do.
print(rsa_getkey())
p: 83
q: 23
n: 1909
e: 19
d: 1899
phi: 1804
The above sets the key values, in which the public key is represented by [n, e] and the private key is represented by [n, d]. This is accomplished with the following code:
function rsa_getkey()
rsa_e = 0
local primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 57, 71, 73, 79, 83, 89, 97}
math.randomseed = os.time()
rsa_p = primes[math.random(5,#primes)]
rsa_q = rsa_p
while rsa_q == rsa_p do
math.randomseed = os.time()
rsa_q = primes[math.random(5,#primes)]
end
rsa_n = rsa_p*rsa_q
rsa_phi = (rsa_p-1)*(rsa_q-1)
while rsa_e == 0 do
local prime = primes[math.random(1,10)]
if rsa_phi%prime > 0 then
rsa_e = prime
end
end
for i = 2, rsa_phi/2 do
if ((i*rsa_phi)+1)%rsa_e == 0 then
rsa_d = ((i*rsa_phi)+1)/rsa_e
break
end
end
return "p: ",rsa_p,"\nq: ",rsa_q,"\nn: ",rsa_n,"\ne: ",rsa_e,"\nd: ",rsa_d,"\nphi: ",rsa_phi,"\n"
end
After the keys have been determined, you can encrypt the message. In order to convert plain text ("Hello world") to a numeric system, I've created a function that isn't 100% complete, but works in the most basic form:
print(rsa_plaintext("Hello_world"))
1740474750625850534739
The following function is how that message is determined:
function rsa_plaintext(x)
local alphanum = {A=10, B=11, C=12, D=13, E=14, F=15, G=16, H=17, I=18, J=19, K=20, L=21, M=22, N=23, O=24, P=25, Q=26, R=27, S=28, T=29, U=30, V=31, W=32, X=33, Y=34, Z=35, a=36, b=37, c=38, d=39, e=40, f=41, g=42, h=43, i=44, j=45, k=46, l=47, m=48, n=49, o=50, p=51, q=52, r=53, s=54, t=55, u=56, v=57, w=58, x=59, y=60, z=61, _=62}
rsa_cipher = ""
for i = 1, #x do
local s = x:sub(i,i)
rsa_cipher = rsa_cipher .. alphanum[s]
end
return rsa_cipher
end
Lastly, in order to make this much more manageable, I have to break it down into segments. In an effort to save time and code, I've combined the actual encryption with the conversion from plaintext to numeric format to encryption, though I've added decryption for debugging purposes. The code also accounts for affixing 0's to message to ensure 4 digits in each grouping. This is where my problem comes in; the Msg and Decrypted should be identical.
print(rsa_group("Hello world"))
Msg: 1740
Encrypted: 1560
Decrypted: 1740
Msg: 4747
Encrypted: 795
Decrypted: 929
Msg: 5062
Encrypted: 1659
Decrypted: 1244
Msg: 5850
Encrypted: 441
Decrypted: 123
Msg: 5347
Encrypted: 429
Decrypted: 1529
Msg: 3900
Encrypted: 1244
Decrypted: 82
This is done with the following two functions:
function rsa_group(str)
local cipher = {}
local str = rsa_plaintext(str:gsub(" ","_"))
local len = #str
local fillin = ""
if len%4 ~= 0 then
fillin = string.rep(0,(4-len%4))
end
str = str..fillin
for i = 1, #str, 4 do
local s,e = i, i+3
local part = str:sub(s,e)
print(rsa_encrypt(part))
end
end
function rsa_encrypt(msg)
bnrsa_e = BigNum.new(rsa_e)
bnrsa_n = BigNum.new(rsa_n)
bnmsg = BigNum.new(msg)
result = 0
quo = BigNum.new()
rsa_c = BigNum.new()
result = BigNum.pow(bnmsg, bnrsa_e)
BigNum.div(result, bnrsa_n, quo, rsa_c)
bnrsa_c = BigNum.new(rsa_c)
bnrsa_d = BigNum.new(rsa_d)
result = 0
quo = BigNum.new()
rsa_C = BigNum.new()
result = BigNum.pow(bnrsa_c, bnrsa_d)
BigNum.div(result, bnrsa_n, quo, rsa_C)
return "Msg:",msg,"\nEncrypted:",rsa_c,"\nDecrypted:",rsa_C,"\n"
end
Now, I know this is a long question, and there are many components to the problem itself. I'm just at a loss how to figure out where my problem lies. Is there something I'm missing? A fresh set of eyes might be my solution.

Answer 1

Upon closer examination it looks like the message M has to be less than the product n of your two primes. In your above test cases, all the messages except the first failed to decrypt properly because they're greater than n = 1909.
For example, consider where M just exceeded n = 1909:
Msg: 1910
Encrypted: 1
Decrypted: 1
Msg: 1911
Encrypted: 1222
Decrypted: 2
Msg: 1912
Encrypted: 1179
Decrypted: 3
In a real-world example, n is of course significantly larger and so this problem is much less likely to arise.