RSA with p = q (Repeated Prime Factor)

Updated 19 Sep 2026

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What it is
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  If RSA key generation accidentally reuses the same prime for both factors ( $p = q$ ), the modulus becomes a perfect square, $N = p^2$ — trivially factored by taking an ordinary integer square root, no specialized factoring algorithm needed at all.
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When to apply
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  Any hint that a "unique" large prime was generated once and reused (source code explicitly doing this, or a modulus that happens to be a perfect square) is worth checking with isqrt(N)**2 == N before reaching for any real factoring tool.
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Math
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  $N = p \cdot p = p^2 \Rightarrow p = \sqrt N$ exactly. $\phi(N) = \phi(p^2) = p(p-1)$ — Euler's totient of a prime power, not $(p-1)^2$ .
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Worked example
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  Let $p=7$ , $N=49$ , $e=5$ , and $m=2$ . Since $\varphi(49)=42$ and $\gcd(5,42)=1$ , $d=17$ . Encryption gives $c=2^5\bmod49=32$ and $c^{17}\bmod49=2$ . The standard exponent argument applies to units modulo $p^2$ ; messages divisible by $p$ require separate care because exponentiation need not be a permutation on those non-units.

Python

from math import isqrt
p = isqrt(N)
assert p * p == N
phi = p * (p - 1)
d = pow(e, -1, phi)