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I'm learning about the Pollard rho factorization algorithms and I couldn't understand one point in the derivation of the algorithm.

Suppose that we want to factor a number $n$ and $p$ is the least prime factor of $n$. There is a function $F$ defined as $F(x) =x^2 + 1 \mod n$ and the following relationship is correct $$ F^{(j)}(s)=F^{(k)}(s) \mod p $$ How is it possible to show that $\gcd(F^{(j)}(s)-F^{(k)}(s),n)$ is divisible by $p$?

Konstantin
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    That's pretty much immediate. $p \mid F^{(j)}(s) - F^{(k)}(s)$ is an assumption. – Daniel Fischer Nov 12 '17 at 19:43
  • @Daniel Fischer Yes, the expression you wrote is immediate (by compatibility with translation) and $F^{(j)}(s)-F^{(k)}(s)$ and $n$ are definitely divisible by $p$. But why their $\gcd$ is divisible by $p$? – Konstantin Nov 12 '17 at 19:56
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    $p$ is a common divisor. And by definition of the greatest common divisor, every common divisor divides it. – Daniel Fischer Nov 12 '17 at 20:00
  • @Daniel Fischer, Got it, thank you very much! https://math.stackexchange.com/questions/198787/prove-that-if-c-is-a-common-divisor-of-a-and-b-then-c-divides-the-gcd-of-a-and-b – Konstantin Nov 12 '17 at 20:40

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