Crypto noob here, I am attempting to do this programming challenge. I do not have the secret key that is used to decrypt the message. However, the key is small enough for a brute force approach. I am attempting to write a function that will solve for the secret key with other variables used in the process.

She tells Bob values of $p$ and $pe \equiv p^e \pmod{n}$ (p raised to power of e mod n) as her public key. Meanwhile the value e remains her secret key (there is no easy way to calculate it from $p$ and $p^e$).

I have $pe$, $p$, and $n$. Here is some code written in java.

static int findE(int n, int p, int pe) {
    for(int e = 0; e < n; e++) {
        if(modular_pow(p, e, n) == pe)
            return e;
    return -1;

static int modular_pow(int base, int exponent, int modulus) {
    int c = 1;

    for(int i = 0; i < exponent; i++) {
        c = (c * base) % modulus;
    return c;

I calculate $p^e \pmod{n}$ for all possible values, but my "findE" function always returns -1. I have tested my modular_pow function and am certain that is not the problem. Maybe I am misunderstanding the instructions. Thanks for any help!

  • $\begingroup$ The code looks as if it should work (albeit quite inefficiently). Maybe you made a mistake somewhere else? What are the arguments you're calling findE with? $\endgroup$
    – yyyyyyy
    Jul 12, 2015 at 20:42
  • $\begingroup$ I just added a System.out.println() on the parameters of my function, here is the resulting output: n: 1000133 p: 372453 pe: 464079 $\endgroup$ Jul 12, 2015 at 20:50
  • $\begingroup$ Ah, here's an explanation: The multiplication c * base overflows since both inputs may be as large as $n$, which is about 20 bits. Hence the product becomes too large for an int and gets truncated. Use arbitrary-precision integers to make it work. $\endgroup$
    – yyyyyyy
    Jul 12, 2015 at 21:04
  • 2
    $\begingroup$ You are right! I didn't think about that one calculation of c * base. Well done, that solved the problem. Arbitrary-precision integers did the trick. $\endgroup$ Jul 12, 2015 at 21:20
  • $\begingroup$ If you know the basic elements all fit within the length of int, then you can use long instead, so that you can do the multiplication steps. Arbitrary-length integers are preferable of course, but if this is a one-time assignment, long would be sufficient and easier to handle. $\endgroup$
    – tylo
    Jul 13, 2015 at 10:42

1 Answer 1


Given $p$ and $p_e \equiv p^e \pmod{n}$. You need to calculate $p_e$. Propose $e$ is small enough to brute-force. The simplest way would be to calcute $p^e \pmod{n}$ for values $e \in \{1,2, ..., n-1\}$ until you find $p_e$.

So all you need would be something like this:

static int findE(int n, int p, int pe) {
    int base = p;
    for(int e = 2; e < n; e++) { // also it suffices to start at e=2
        p = p * base % n;
        if (p == pe) return e;
    return -1;

Even though you are doing to much calculation (for every value of $e$ you recalculate all values of $p^{e-1}$), right now i do not see why your code would'nt work. Update: forget all i said about modular exponentation, that was bullsh*t.

  • $\begingroup$ my modular_pow function computes using modular exponentiation because n can be of around 100,000 which means I have to use a memory efficient method to perform the calculations. $\endgroup$ Jul 12, 2015 at 20:38
  • $\begingroup$ Ahh I see how your way is better now! You multiply it and then use modulus for each iteration and that way it never gets extremely large. Thank you! $\endgroup$ Jul 12, 2015 at 20:47
  • $\begingroup$ I simply calculate each $p^i$ for all possible $i$ continously. You calculate $p^1 = p$, $p^2 = p * p$, $p^3 = p * p * p$, ..., and each time to start again with $p$ instead with the value you calculated in the previous iteration. Since you calculated $p^{i-1}$ in the previous round, there is no need to start all over again. $\endgroup$
    – Fleeep
    Jul 12, 2015 at 21:00
  • $\begingroup$ Yes, much better than my way. Also, I copy-pasted your method and it still returned -1 with the arguments n: 1000133 p: 372453 pe: 464079 $\endgroup$ Jul 12, 2015 at 21:01
  • 1
    $\begingroup$ "i just did a quick run in python with my code..." Python has arbitrary-length integers by default, Java doesn't, which seems to be the problem. $\endgroup$
    – tylo
    Jul 13, 2015 at 10:47

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