The RSA modulus is the product of two $2048$-bit primes.
And the two Public Exponents are both $16$-bit.
I also got the difference between two Private Exponents $\left | d_1-d_2 \right |.$
Is there any way to factorize the Modulus $N$?
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$\begingroup$ What is the origion of this Q? How much this difference? $\endgroup$– kelalakaOct 9, 2022 at 21:51
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$\begingroup$ diff/n is about 3n/5,I thought it looks like some special trick in Cryptanalysis of RSA with two decryption exponents $\endgroup$– MancOct 9, 2022 at 22:00
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$\begingroup$ you mean absolute difference not difference divided by n clearly $\endgroup$– kodluOct 9, 2022 at 22:32
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$\begingroup$ So you know $N,e_1,e_2, |d_1-d_2|$ with small $e_1$ and $e_2$, and want to factor $N$. Hint: write a relation that must exist between $e_1$ and $d_1$, same between $e_2$ and $d_2$. If these contain$\bmod$, apply the definition of that to remove it. And proceed to adapt the usual method to factor $N$ given $e$ and $d$. $\endgroup$– fgrieu ♦Oct 10, 2022 at 5:47
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