# Pragyan CTF 2019 "Easy RSA" writeup
## Description
Deeraj is experimenting with RSA. He is given the necessary RSA parameters. He realizes something is off, but doesn't know what. Can you help him figure it out?
[parameters.txt](parameters.txt)
## Solution
First, I noticed that the `e` parameter (public exponent) is very large. In this case, the secret key `d` can be calculated by using Wiener's attack.
> parameter.txt を見ると、RSA暗号に使われる e は一般的に 65537(= 2^16 + 1) であるのに対し、かなり大きな値になっている。この時、Wiener's Attack が有効であると考えられる。Wiener's Attack は 秘密鍵 d が十分小さいときに、公開鍵から秘密鍵を復号できる。この場合は e が大きいために d が相対的に小さくなることを利用している。
I used an implementation of Wiener's attack on Github.  [rsa-wiener-attack](https://github.com/pablocelayes/rsa-wiener-attack)
> Wiener's Attack の [実装例](https://github.com/pablocelayes/rsa-wiener-attack) がGitHubにあるので、これを用いて秘密鍵 d を得る。
`d = 12978409760901509356642421072925801006324287746872153539187221529835976408177`
Let `m` be the plaintext, m ≡ c^d mod n.`m` is the flag, but we have to convert m(decimal) into a string.
(m(Dec) -> m(Hex) -> m(String))
> 平文 m は m ≡ c^d mod n で表されるので、m を計算し文字列へと変換すると Flag が獲得できる。
```pythone = 217356749319385698521929657544628507680950813122965981036139317973675569442588326220293299168756490163223201593446006249622787212268918299733683908813777695992195006830244088685311059537057855442978678020950265617092637544349098729925492477391076560770615398034890984685084288600014953201593750327846808762513n = 413514550275673527863957027545525175432824699510881864021105557583918890022061739148026915990124447164572528944722263717357237476264481036272236727160588284145055425035045871562541038353702292714978768468806464985590036061328334595717970895975121788928626837881214128786266719801269965024179019247618967408217c = 337907824405966440030495671003069758278111764297629248609638912154235544001123799434176915113308593275372838266739188034566867280295804636556069233774555055521212823481663542294565892061947925909547184805760988117713501561339405677394457210062631040728412334490054091265643226842490973415231820626551757008360d = 12978409760901509356642421072925801006324287746872153539187221529835976408177
m = hex(pow(c,d,n))[2:]
flg = ''for a, b in zip(m[::2], m[1::2]):    flg += chr(int(a+b, 16))
print(flg)
```
Flag : `pctf{Sup3r_st4nd4rd_W31n3r_4tt4ck}`