[HackingWeek 2015] [Crypto3] Write-up

Description

Alice wants to use the ECDSA signature algorithm.
For this, she chooses to use the elliptic curve equation E where y2 = x3 – x, defined on the finite field with q elements, Fq, with q = 134747661567386867366256408824228742802669457 .

She chose P as a base point of order N = 2902021510595963727029 with:
P = (18185174461194872234733581786593019886770620, 74952280828346465277451545812645059041440154)

Then, she chooses as secret signature key a random integer d with 1<d<N.
The public key is the point Q=d⊗P where ⊗ is the scalar multiplication on the elliptic curve E.

The value of Q is:
Q=(76468233972358960368422190121977870066985660, 33884872380845276447083435959215308764231090)

The validation key is given by the value of d.

Continue reading [HackingWeek 2015] [Crypto3] Write-up

[HackingWeek 2015] Back on the event

The HackingWeek is now over !

Good challenges overall, but the staff lacked organization.
Although with a single administrator to manage everything, it shouldn’t be easy either!
Moreover we would have preferred a page announcing the challenges updates (and especially their modifications) rather than be surprised when someone from the team found out.

In terms of ranking, the first days we managed to keep a place in the Top 10 without too much difficulty.
After a week and 18/20 solved challenges, we were ranked first.
The days that followed, our position has decreased to finally be ranked 5th out of 106, with 19/22 solved challenges.
This is not so bad after all, especially since it was the first CTF for most of us!

It’s clear that our team unfortunately does not have enough skill in reverse/exploit 🙂
(For the next CTF we will revise these themes, because yes -to general demand- we will do it again!)

Anyway, we’re just finishing the writeups to share them.

Stay tuned !

Crypto: crypto1 – crypto2crypto3crypto4
Exploit: exploit1 – exploit2 – exploit3exploit4exploit5
Forensic: forensic1forensic2forensic3forensic4
Network: net1net2net3 – net4
Reverse: reverse1 – reverse2