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As for the algorithm, you have two choices: The quadratic sieve and the general number field sieve. If you want to implement the factorization yourself you want to use the quadratic sieve, as a good implementation of the GNFS is really hard. However, if you're willing to use pre-made tools using GGNFS and msieve (together) is your best option. Msieve ...

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What is the largest integer that can be factored in 5 hours with a normal computer? That depends on a number of points. What do you consider a "normal" computer? A 2015 Gamer-PC with a high-end quad (4) or octo (8) core CPU and more than one graphics card? A 2010 Gamer-PC with a back-then modern CPU and a modern GPU? A 2012 Multimedia-PC with a ...

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Summary: to a considerable degree, more computers speed up factorization of a given integer; but the expected time decreases significantly slower than the inverse of the number of computers used: we are in the area of sub-linear speedup. Some high-performance (but not the best) factorization algorithms, in particular ECM, enjoy near-linear speedup with ...

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The original poster clarified that: the application of RSA is signature (not encryption as originally stated); at least one signature $s$, the value of $N$, and $e=3$ are given; the signature is by signing an MD5 hash of a message, and a hash of the message matching the signature $s$ is also a given. The first, low-effort thing to do with the givens is ...

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The angel on my other shoulder says the following: Every polynomial-form approach to finding the factors of a numbers ultimately boils down to a search for integer solutions to the radical expression: $$\epsilon = \varepsilon \pm{} \sqrt{\xi^2 \pm N}$$ and so the use of quadratic or cubic spatial interpolation and filtering will inevitably lead to a ...

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According to Wolberg monotonic interpolation techniques exist which can reduce the oscillation seen in the graph mentioned in the the question's postscript. I'm not going to discuss Wolberg's method, but I will use background information from his paper to compute a formula for the original graph. Depending on how it all goes, I leave it to the reader to ...

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