Is it possible to calculate the private exponent when only the RSA public key (e;N) = (9292162750094637473537; 13029506445953503759481) is given?
$\begingroup$
$\endgroup$
-
$\begingroup$ That would only be possible for definitions of RSA that uniquely define the private exponent, such as FIPS 186-4; not for definitions of RSA that allow several private exponents, such as PKCS#1. $\endgroup$ – fgrieu Nov 27 '16 at 4:22
$\begingroup$
$\endgroup$
Yes, since $N=p \times q$, so we have $13029506445953503759481 = 109217267039×119298960679$, also the private and public exponent $e$ and $d$ are related by the equation $ed\equiv1\text{ mod }\phi(N)\equiv1\text{ mod }((109217267039-1)×(119298960679-1)) $. You should then be able to find $d$, the private exponent.
-
-
2$\begingroup$ @ArsalanAmjid probably by handing the modulus to a factoring tool? $\endgroup$ – SEJPM♦ Nov 26 '16 at 17:47
-
1$\begingroup$ @Arsalan Wolfram Alpha factored it in about 3 seconds. Note that in real life, N (and hence p and q) would be much, much larger... $\endgroup$ – Dan Nov 26 '16 at 20:11
-
1$\begingroup$ Yes, @Dan is right, now the standard for the modulus $N$ is at least $1024$ bits. $\endgroup$ – freak_warrior Nov 27 '16 at 3:54
