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I have given n, e, c, and 2d+phi(n). any idea how to proceed to solve that RSA. for values click on this:- https://pastebin.com/mdeSdfzD

Mark
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    What have you tried, where are you stuck? This question shows absolutely no effort. – Maeher Feb 17 '18 at 03:40
  • I have tried the same thing which is given in below comment. ed = 1+k*phi(n). but there is another variable k, after which I am not able to understand what to do next. – Mark Feb 17 '18 at 13:19
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    he's cheating on easyctf. https://easyctf.com – Kevin Feb 17 '18 at 21:46

1 Answers1

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p = 179625470269984575211291664517722616039787182426068867084101295486932993749787129633737153741496747351573119936946306867775629661781307731728093871384501392452057213703678422306940017327268991666011108647323837661852433288372793079254503318017584711837671484712835436166606958277493346384859583473813750793903

q = 165951900395573838473861177483977329257584913855055564650331190368208599646122670409920222286702764849767512840736219293650470102124983099032047151432801085844142882059163396523057801419384753318562498454796480286799673413582017708953121180295255267080043853111452400359241673145348291220080289048917975182701

Hint: $(2d + \phi(n))\cdot e \equiv 2 \pmod{\phi(n)} \implies (2d + \phi(n))\cdot e - 2$ is a multiple of $\phi(n)$

user94293
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    How you get p and q, Can you explain more, I am not able to understand it, it's important for me – Mark Feb 17 '18 at 13:12
  • See also https://crypto.stackexchange.com/questions/6361 – user94293 Feb 18 '18 at 22:15
  • This doesn't make sense-- his 2d+phi(n) multiplied by e in his example is a multi-hundred digit number. And 2 mod (number > 2) always equals 2. – Aaron Esau Feb 19 '18 at 19:03
  • @Arin: Above notation is for congruence. It means $[(2d+\phi(n))\cdot e] \bmod \phi(N) = 2 \bmod \phi(N)$. – user94293 Feb 19 '18 at 19:16
  • @Mark And as soon as you know a multiple of $\phi(N)$ you can factor $N$ using the approach as when you know the public and private key but not the primes. – SEJPM Feb 21 '18 at 10:03