rsa-59 = 71641520761751435455133616475667090434063332228247871795429 (59 digits, factors p=200...437 and q=357...017 have both 30 digits)

rsa-79 = 7293469445285646172092483905177589838606665884410340391954917800303813280275279 (79 digits, factors p=848...977 and q=859...727 have respectively 39 and 40 digits)

rsa-99 = 256724393281137036243618548169692747168133997830674574560564321074494892576105743931776484232708881 (99 digits, factors p=486...193 and q=527...217 have respectively 49 and 50 digits)

rsa-119 = 55519750778717908277109380212290093527311630068956900635648324635249028602517209502369255464843035183207399415841257091 (119 digits, factors p=106...387 and q=520...393 have both 60 digits)

Once you have solved the above challenges, to really convince me you have found an efficient factoring algorithm, please do the following:

- pick up a large unfactored publicly known integer, say
`N`(RSA-1024 should be enough to convince me and many other people); - from the factorization
`N=pq`you have, deduce the private key`d = 1/e mod (p-1)(q-1)`corresponding to the public key`e=65537`; - compute
`c = 3`;^{d}mod N - send the integer
`c`to me (or publish it on some web page); - on my side, I will compute
`m = c`and check that^{65537}mod N`m=3`.

- you cannot cheat since for that you need to extract an
`e`-th root of`N`; this is known as the RSA problem; - I cannot deduce the factorization of
`N`from the value of`c`.

Thanks to my colleagues Guillaume Hanrot, Pierrick Gaudry and Emmanuel Thomé, who helped me improving this page.