So I've been learning about RSA for quite a while (mainly by playing around in CTF competitions) and I came across an interesting problem.

The other day I was looking to create a challenge in which I needed $\phi(n)$ to be a perfect square, but under the condition that $p \neq q$.

For very small bitlengths (such as $<= 20$ bits) a naive approach such as the following will be enough to find such primes:

from Crypto.Util.number import *
from gmpy2 import iroot

p, q = getPrime(15), getPrime(15)

while p == q or not iroot((p - 1) * (q - 1), 2)[1]:
    p, q = getPrime(15), getPrime(15)

print(p, q)

But if you're looking for 512 and 1024 bit numbers, this is like searching for a needle in a haystack.

And so, my question to you is, is there some efficient approach to do this?

I've looked all over Google, but nothing of this sort pops up :/

  • 1
    $\begingroup$ Generate a perfect square torsion of small primes, randomly assign the primes to $p-1$ and $q-1$, check that $p$ and $q$ is prime with a fast primality test like Rabin Miller. $\endgroup$
    – kelalaka
    Dec 6, 2023 at 17:30
  • $\begingroup$ And, if you need a $pq$ that is hard to factor, you can use several not-very-small primes to create your square; it'll be more work (there might not be set that makes both $p$ and $q$ prime, and so you'll need to go through several sets), but it is still feasible. $\endgroup$
    – poncho
    Dec 6, 2023 at 17:45
  • 1
    $\begingroup$ @kelalaka: you should make your comment the answer $\endgroup$
    – poncho
    Dec 6, 2023 at 17:46
  • $\begingroup$ @kelalaka So essentially find small primes such that their product is a perfect square with 2*(desired_bitlength) bits and try to recover p and q such that p-1 and q-1 give us the same product? Wouldn't that need to check a lot of permutations when it comes to splitting the list of primes? $\endgroup$
    – Anonymous
    Dec 6, 2023 at 17:49
  • $\begingroup$ @poncho fgrieu's answer is better and I my answer will be half since the probability of hitting a prime is need.. $\endgroup$
    – kelalaka
    Dec 6, 2023 at 22:39

2 Answers 2


We restrict to RSA modulus $n=p\,q$ with $p$ and $q$ distinct primes, thus $\phi(n)=(p-1)(q-1)$.

For any given $(p,q)$ such that $\phi(n)$ is a square, there exists $(a,b,g)$ with $p=a^2g+1$, $q=b^2g+1$. We have $\sqrt{\phi(n)}=a\,b\,g$, and $a\ne b$. Further that $(a,b,g)$ is unique if we add $g=\gcd(p-1,q-1)$, and then $a$ and $b$ are coprime, and $g$ is even.

Thus to generate the desired $p$ and $q$:

  • pick random even $g$ of roughly $1/4$ the desired bit size of $n$.
  • pick random $a$ of roughly $1/8$ the desired bit size of $n$, until $p=a^2g+1$ is prime.
  • pick random $b$ of roughly $1/8$ the desired bit size of $n$, until $\gcd(a,b)=1$ (optional, see below), and $q=b^2g+1$ is prime, and $b\ne a$.

We can remove the requirement that $\gcd(a,b)=1$ in the generation of $b$, and for a fixed size of $g$ and $a$ that increases the number of $n$ that can be generated, which can be viewed as beneficial. However I fear this tends to increase the multiplicity of some primes in the factorization of $\sqrt{\phi(n)}$, which could be bad.

If it was not for the prescribed sizes, this procedure could generate any $n$ such that $\phi(n)$ is a square. I'm not sure about the size of $g$ that will make $n$ the hardest to factor.

An initial version of my answer, and the algorithm there, use $2c$ where there is now $g$. Also I restricted to $a$ and $b$ both odd, and incorrectly stated this is necessary when really it about halves the number of possible $n$.


Thank you so much @fgrieu for the detailed and understandable answer. I quickly implemented it in python for anyone else interested in this:

from Crypto.Util.number import *
from gmpy2 import iroot
from random import getrandbits

def generate(bitlength):
    c_bitlength = bitlength // 4 # c is meant to be 1/4th of the modulus bitsize
    a_bitlength = bitlength // 8 # a is meant to be 1/8th of the modulus bitsize

    while True:
        # Step 1: pick random c
        c = getrandbits(c_bitlength)

        while True:
            # Step 2: pick random odd a until p is prime
            a = getrandbits(a_bitlength)
            p_candidate = 2 * a**2 * c + 1

            if isPrime(p_candidate):

        while True:
            # Step 3: pick random odd b until q is prime and b != a
            b = getrandbits(a_bitlength)

            if b != a:
                q_candidate = 2 * b**2 * c + 1

                if isPrime(q_candidate):
                    phi_n = (p_candidate - 1) * (q_candidate - 1)

                    # Check if phi is a perfect square
                    if iroot(phi_n, 2)[1]:
                        return p_candidate, q_candidate

bitlength = 512
p, q = generate(bitlength)
print("p:", p)
print("q:", q)

Thanks again!

  • 1
    $\begingroup$ Note that we are not a coding site and don't want code-only answers. We use the code to demonstrate the theory. Normally, this should be posted on GitHub and similar sites and commented on under the answer. $\endgroup$
    – kelalaka
    Dec 7, 2023 at 9:33

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