0
$\begingroup$

I've got two 1024 bits prime $p$,$q$,and $n$ = $p$ * $q$. now I know the result of $ c^{p} \quad mod \quad n = x$,also the value of c is given, I wonder if it is possible to factorize $n$.

$\endgroup$
2
$\begingroup$

From Fermat's little theorem we know $$ a^p \equiv a \pmod{p}\,. $$ Applying this to the present problem, $c^p \equiv c \equiv x \pmod{p}$, and thus $p = \gcd(x - c, n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.