No Padding, No Problem

Oracles can be your best friend, they will decrypt anything, except the flag's ciphertext. How will you break it? Connect with nc 10333

Upon connecting, we get the values of NN and ee as well as the encrypted ciphertext cc that represents the flag. We then have a decryption oracle, which can decrypt anything except for the flag.

Note that the ciphertext is decrypted as follows:

mcdmodNm \equiv c^d \mod N

If we ask to decrypt c-c instead, we get

m(c)dcdmodNm \equiv (-c)^d \equiv -c^d \mod N

Note the last congruence is because dd is odd, so (1)d=1(-1)^d = -1.

This means that if we pass in the negative of cc, we can get the negative of the decryption!

N = 64225632402784743608151428388331019007158039700441403609620876723228303996217136829769322251101831115510439457268097599588978823846061420515078072743333076016253031234729517071419809456539618743788851473244412318432363995783182914809195026673348987512316519371501063936603604905070428868194818209957885002651
R = IntegerModRing(N) 
c = R(23961525860638788006091919862301366730415613260613078904461027043559403510831473561860834624403033454974614369313881141911510211211764847671996788759608002057996932820692709010900418723347410147858586280735791816478632919784849715797867137711835451159040091442311708166252069010315360215005284477472628144578)

# send it back, get result
negative_m = R(64225632402784743608151428388331019007158039700441403609620876723228303996217136829769322251101831115510439457268097599588978823846061420515078072743333076016253031234729517071419809456249343713593001433770955700064908110713217165957916949916605267065613204854099704669280835867601177422810391570120236404254)

# picoCTF{m4yb3_Th0se_m3s54g3s_4r3_difurrent_1772735}

There are other ways to do it too - you could calculate 265537modN2^{65537} \mod N and multiply cc by that, which would yield you 2c2c after decryption, and you'd just need to halve it, as described in this writeup.

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