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Unable to factorise the modulus from public key to extract a private key

in flag
pee2pee

I have a public key but struggling to get the modulus from it

alan@WW031779:/mnt/c/Users/alan/Downloads/rsatool$ openssl rsa -noout -text -inform PEM -in lol.pem -pubin
RSA Public-Key: (2048 bit)
Modulus:
    00:e6:23:97:28:84:b1:f4:d7:22:bd:d5:ee:5b:eb:
    84:cb:84:76:0c:2e:d0:ff:af:d9:3c:d6:03:0f:b2:
    0d:79:30:90:3b:d1:73:1d:c7:4c:95:4a:23:07:53:
    03:df:d7:1b:88:5c:d6:6e:98:5b:f7:59:ed:17:a9:
    85:f7:e7:d8:37:c8:57:bd:31:a1:47:d7:4d:a2:61:
    49:28:58:fa:5f:cf:b8:92:30:87:8e:f4:ff:fc:ff:
    92:fc:29:29:89:32:64:54:af:b5:1b:b7:ab:25:3f:
    ef:d5:b3:57:bf:83:a6:39:f1:53:20:4a:fc:56:28:
    f3:e0:20:22:c6:94:9d:c2:3c:b1:9d:2f:d6:39:b6:
    d5:98:7a:c3:32:a0:1d:d2:3b:43:7a:67:77:bb:96:
    7f:80:e5:22:e9:41:e5:f9:72:16:0a:ed:55:6d:b7:
    39:39:19:80:64:22:ae:1a:7d:c9:b1:99:96:fd:b7:
    b2:91:41:47:2d:68:03:df:f4:2a:71:3d:b5:7a:c0:
    78:fc:a4:8d:1a:68:61:42:3d:e3:a1:2e:d9:cf:af:
    b8:31:e5:d6:9b:92:d7:19:63:d0:23:22:8c:26:12:
    ea:33:4a:65:2c:46:12:1f:50:5d:1b:5a:55:12:24:
    c6:9f:c8:23:9c:fe:10:93:de:68:09:5f:71:53:15:
    96:67
Exponent: 3 (0x3)

This gives

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

In decimal it is equal to

29052360453120059177701146498207729611014362120841772147885284668310294675407700581246333337318872050600353022438909391852076208405990507154764842795064455368228014381969783955360165426546947312973195061115837228105648770122442650123819968683831588775039837617788817831554836487051931001049480790287468125246758818911220414888048673899462271009956700067150701189256017793349102117503912782889345559816174845605183913828898737756645848010661322897081850561427949550036638510279173557403134806365178654334553002357480355906235714208451535185647256346503450896572487047615057007598805977277186223884121839444217172432487

However on factodb.com it doesn't factorise. I was trying to follow https://0x90r00t.com/2015/09/20/ekoparty-pre-ctf-2015-cry100-rsa-2070-write-up/ but now stuck as where to go.

Any advice?

95
0 + 0
rsa
poncho avatar
my flag
poncho
Factorizing a well-chosen 2048 bit modulus is infeasible at the current time. Unless it is known that this modulus was selected to be easy to factor (say, in a CTF challenge), you are quite unlikely to succeed.
3
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in flag
pee2pee
It is a CTF style lab at work so I imagine so but not sure where to start...
0
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poncho avatar
my flag
poncho
Fermat factorization works if the two factors were chosen to be quite close together (which they may for a CTF challenge). Alternatively, one of the factors may be quite small (perhaps 100 bits); in that case, downloading an ECM factorization tool and having that run for a couple of hours may work. I'd try both (because they both work on different weaknesses that the CTF authors may have selected)
3
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Daniel S avatar
ru flag
Daniel S
HINT: The fact that the encryption exponent is 3 suggests that it might not be necessary to factor the modulus to capture the flag.
3
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Don Freecs avatar
sz flag
Don Freecs
Coppersmith's attack
0
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