20 lines
1.9 KiB
Python
20 lines
1.9 KiB
Python
from flag import flag
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from Crypto.Util.number import *
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# Is it possible to factor 2048 bits RSA????
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p = getStrongPrime(1024)
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q = getStrongPrime(1024)
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n = p * q
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e = 65537
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message = bytes_to_long(flag)
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cipher = pow(message, e, n)
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hint = (p >> 256) << 256
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print('hint =', hint)
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print('n = ', n)
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print('cipher = ', cipher)
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'''
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hint = 170021659067442061523756541191993539248601165314725269340063885026090819928194021233489114706002869522375199940625136026822330472394573193982550217044532146962203411567638298486561460035048451998397915884861905155825573014911356931040204843324084786281107112092211904426426572148513199575481399847133231710208
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n = 26572534665683235245636883276068606612246011683144885473463339711820331993493357279285848391008791322234296406608539840620252625886064149394162998948970494176467317183311068211754639508369506106626167584042532674048767021259563143175030276397301094824563794971045331643148544039656526615133318494601670688100830427083371422333448416345157657081894568628680132969792458721329931299527818220689347292610785587787837487603664368727957712489422350366840751269348100479824826707549651813948243868760028344290958809882314383956517334878930614926458454637703779278161529934602431923391507121553686269628353968327172758558101
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cipher = 6757305476823187630788813544175078002034639295382430908622959031152682951668576844271799811383004513343837564381952708716729553254090143798506889925295074373036913929804491380602451212049174613264171604463300669869149837525862922090696673791849544763152707468306110063163234629066598567091839169641464962113876354673950799668460699078314705550375034690000052853842247971910604379818131880965835942702654798359475442300498917978942357402126717495842970216610908374556549274105304987976888600967085331993375097585293113427600688155704013684617306744103329525203776458981287207246516848585805560812026318668317699834254
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''' |