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if you know that $n=3598057$ is product of two prime distinct numbers and also that $20779$ is root from $1$ mod $n$, find factor integer for $n$.

From wiki I expected that $$20779^2 \equiv 1 \mod n$$ But I checked in calculator that

$$20779^2 \equiv 0 \mod n$$ So I am not sure if I have understood this term correctly.

Update

I tried to delete this question on phone (During writing this post I didn't have access to pc). But now, when there is a answer, I decided to mark as the best contributor and left this task for future readers. Exactly, my phone hasn't got a proper precision for numbers like that.

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$20779^2 \equiv 1 \pmod {3598057}$ so

$20779^2 -1 \equiv 0 \pmod{3598057}$

$(20779-1)(20779+1) \equiv 0 \pmod {3598057}$

SO $3598057| 20778*20780$ so the prime factors of $3598057$ are among the prime factors of $20778$ and $20780$.

So $20778 = 6*3463$ and $20780 = 20*1039$.

$6$ and $20$ are clearly relatively prime to $3598057$ so the two prime factors of $3598057$ must be among the prime factors of $3463$ and $1039$.

It's within the realm of reasonability (I guess) to check that $3463$ and $1039$ are prime and that $3463*1039=3598057$.

.... That is, if we were told all those thing. (That $20779$ is square root of $1\pmod{3598057}$ and that $598057$ is the product of two primes.) Not sure how you'd be expected to know that without guidance.

fleablood
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    And if you don't want to factor $20778$ and $20780$, you can compute $\gcd(20778,3598057)$ and $\gcd(20780,3598057)$ using the Euclidean Algorithm, and get factors of $3598057$ that way. – Gerry Myerson Aug 26 '19 at 05:59
  • Well, sure I guess. Easy to see that $20778$ is even and by rule of $3$ divisible by $6$, and that $20780$ is divisible by $20$. But yeah. ... – fleablood Aug 26 '19 at 06:06