RSA exponent size - encryption

Question

I am learning about the RSA algorithm. I perform the algorithm on very small prime numbers and use online Big Integer calculators to perform the encryption and decryption and everything works just fine.
My question is about the size of the exponent we create and when it comes to bigger numbers, it seems infeasible to calculate.
For example, the algorithm starts with picking two prime numbers p and q. You compute n=pxq and then the totient of n. Next you pick a number 'e' such that 1
Then to perform an encryption you take say like the ASCII character 'A' which is 65 and you raise it to the power of e. (65^e)
The online big integer calculator started getting very slow and sluggish (over a minute to calculate) when e was bigger than about 100,000 (6 digits)
My question is then, for the working RSA algorithm, what size (number of digits) number does that algorithm pick?
One thought I had was it was possible the online calculator that I was using was not using the best method for exponents? This is the calculator I am using: http://www.javascripter.net/math/calculators/100digitbigintcalculator.htm

Answer 1

Let's say M is the modulus. So YES, you could first perform intermediate = 65^e, and finally compute intermediate mod M. And of course, intermediate would be a very very very very big integer (if e equals 65537, the decimal representation of intermediate contains 118813 digits!).
BUT, thanks to a very basic modular arithmetic theorem,
(65^e) mod M = ((((65 mod M) * 65) mod M) * 65) mod M [...] (e times)
(the theorem states that in a quotient ring, the n-th power of the class of an element is the class of the n-th power of the element)
As you can see, this does not need any very big integer library, since after each arithmetic product, you use mod M that returns an integer between 0 and M-1. So, you only have to compute arithmetic products of integers less than M.
As an example, here is a simple shell script (bash) that computes 65^65537 mod 991*997. As you can see, no need to get a big number library:
#!/bin/bash
# set RSA parameters
m=65 # message to encode
M=$((991*997)) # modulus (both 991 and 997 are prime numbers)
e=65537 # public exponent (coprime with 990*996, thus compliant with RSA algorithm)
# compute (m^e) mod M
ret=1
for i in {1..$e}
do
ret=$(((ret*m)%M))
done
# display the result
echo $ret
It immediately returns 784933, thus 65^65537 mod 991*997 = 784933
The biggest integer computed with your method of calculus has 118813 digits, but the biggest integer handled with this shell script only has 12 or less digits ((M-1)^2 is made of 12 digits).
According to these explanations, we can now answer your question:
My question is then, for the working RSA algorithm, what size (number of digits) number does that algorithm pick?
With the above explanations, you can see that the maximum number of digits in the decimal representation of integers you have to manipulate is 1+log10((M-1)^2), because you will, at most, compute a product of two integers between 0 and M-1.
Note that 1+log10((M-1)^2) = 1+2.log10(M-1) < 2+2.log10(M) = 2.(1+log10(M)). Also note that 1+log10(M) is the number of digits of M.
Therefore, as a conclusion, this proves that the number of digits your library has to handle correctly is twice the number of digits of the modulus (if you are computing the exponentiation using integer multiplications the way explained here).