I've read some similar posts and the solution which you sugest is Decimal module. But my problem isn't that my code can't count it. I mean once it can and once not. Im interested in RSA and learning encryption and decryption private keys there is simple code:
p=53 #secret
q=59 #secret
n=p*q
k=2 #secret
print(n)
PHI=(p-1)*(q-1) #secret
print(PHI) #secret
#choose a small public exponent, "e" must be odd, have no common divisors with phi (n)
e=3
d=(k*PHI+1)/3 #secret
print(d) #secret
# secret message is 89
c=(89**e)%n
print(c)
x=(c**d)
print(x)
x=x%n
print(x)
"""x=1394**2011
print(x)"""
There is no problem with counting ((89^3) mod n) and there is no problem with last equation (as a comment) 1394^2011= i have number with 6324 digits, but when i try use x=(c^d) mod n i have :
Traceback (most recent call last):
File "RSA.py", line 77, in <module>
x=(c**d)
OverflowError: (34, 'Numerical result out of range')
e is an int, and so 89**e can be arbitrarily large (though I would use pow(89, e, n) so that you don't have to compute the large intermediate value 89**e first).
d, however, is a float, and c**d will also be a float. A float cannot grow arbitrarily large, hence the error.
Since d is supposed to be an int, use floored division.
d = (k*PHI + 1) // 3