added official hacktheboo2024 writeups

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/Makefile

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+default:
+ifdef name
+	@cd challenge; \
+		python3 source.py; \
+		mkdir crypto_$(name); \
+		cp source.py output.txt ./crypto_$(name); \
+		cp output.txt ../htb/; \
+		mv ./crypto_$(name) ../release/;
+
+	@cd release; \
+		zip -9 -r ./crypto_$(name).zip ./crypto_$(name); \
+		unzip -l ./crypto_$(name).zip;
+
+	@echo [+] Challenge was built successfully.
+else
+	@echo [-] Please define the challenge name. For example, \"make name=cool_chall_name\"
+endif
+
+flag:
+	@echo [+] Flag : $$(cd challenge; python3 -c 'print(open("flag.txt").read())')
+solver:
+	@echo [+] PoC : $$(cd htb ; sage -python3 solver.py)
+	@find . -name "*.sage.py" -type f -delete
+
+test: clean default flag solver
+
+clean:
+	@rm -rf release/*
+	@rm -rf htb/output.txt
+	@find . -name "*.sage.py" -type f -delete
+	@echo [+] Challenge release deleted successfully.

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/README.md

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+![](assets/images/banner.png)
+
+<img src="assets/images/htb.png" style="zoom: 80%;" align=left /><font size="5">sugar free candies</font>
+
+​	22<sup>th</sup> September 2023 / Document No. D23.102.XX
+
+​	Prepared by : `rasti`
+
+​	Challenge Author(s): `0x50r4`
+
+​	Difficulty: <font color=lightgreen>Very Easy</font>
+
+​	Classification: Official
+
+​		
+
+ 
+
+# Synopsis
+
+- `sugar free candies` is a very easy crypto challege. The flag of this challenge is splitted into three parts. Four relations between these splitted parts are calculated and outputted to an output text file. The players will have to solve a system of three unknowns, given four equations with SageMath, or any other tool of their choice, and recover the flag.
+
+## Description
+
+- For years, strange signals pulsed through the air on the eve of October 31st. Some said it was the voice of an ancient witch, others believed it was a message from something far darker. A cryptic message, scattered in three parts, was intercepted by a daring group of villagers. Legend spoke of a deal made between the witch and a shadowy figure, but the true intent of their secret could only be revealed by those brave enough to decipher it before midnight, when the veil between worlds would thin.
+
+
+
+## Skills Required
+
+- Basic Python source code analysis.
+- Know how to work with equations with multiple variables programmatically.
+
+## Skills Learned
+
+- Solve equations with SageMath.
+
+# Enumeration
+
+In this challenge we are provided with two files.
+
+- `source.py` : It is the main python script that produces the output file.
+- `output.txt` : Contains the data outputted from the source script.
+
+## Analyzing the challenge
+
+The code of the script is not lengthy so let us take a look.
+
+```python
+from Crypto.Util.number import bytes_to_long
+
+FLAG = open("flag.txt", "rb").read()
+
+step = len(FLAG) // 3
+candies = [bytes_to_long(FLAG[i:i+step]) for i in range(0, len(FLAG), step)]
+
+cnd1, cnd2, cnd3 = candies
+
+with open('output.tcnd1t', 'w') as f:
+    f.write(f'v1 = {cnd1**3 + cnd3**2 + cnd2}\n')
+    f.write(f'v2 = {cnd2**3 + cnd1**2 + cnd3}\n')
+    f.write(f'v3 = {cnd3**3 + cnd2**2 + cnd1}\n')
+    f.write(f'v4 = {cnd1 + cnd2 + cnd3}\n')
+```
+
+The steps can be summarised as:
+
+- The challenge reads the flag from `flag.txt`, which is secret and not provided to us
+- It is splitted in three parts, which are denoted as `cnd1`, `cnd2`, and `cnd3` respectively. Each part is then converted into an integer representation.
+- Four relations between the splitted parts are calculated and finally outputted to the output file `output.txt`.
+
+# Solution
+
+Our task is straight forward. We have to solve a system of three unknowns given four equations. Since the number of unknowns is less than the number of equations we are given, it is guaranteed that a solution $(x, y, z) = (x_0, y_0, z_0)$ exists. It should be noted that three equations would also guarantee a unique solution.
+
+However, the problem is that the numbers are large and the equations non-linear so it would be very hard to solve with just pen and paper. We have to use some tool for this purpose. By researching the keywords `python solve equations` with the search engine of your choice, one should stumble upon the `SymPy` library. However, as the SageMath library is more common in CTFs, we will prefer that one. Note that there are also online equation solvers that should work too.
+
+# Exploitation
+
+Let us first load the data from the output file.
+
+```python
+exec(open('output.txt').read())
+```
+
+Let us first write a function that creates and returns three variables, one for each part of the flag. Due to use of `bytes_to_long`, we know they must be integers so we care only about solutions in the set of integers $\mathbb{Z}$. In SageMath, we can define variables using the `var()` function.
+
+```python
+from sage.all import *
+
+def create_variables():
+		x,y,z = var('x,y,z', domain=ZZ)
+    return x,y,z
+```
+
+Then, we move on solving the system. For this purpose, we will utilize SageMath's [function](https://doc.sagemath.org/html/en/reference/calculus/sage/symbolic/relation.html#sage.symbolic.relation.solve) `solve()`. This functions receives a list of equations and the variables we want to solve for. Let us write a function that returns the solutions of the equations.
+
+```python
+def solve_system(x, y, z, v1, v2, v3, v4):
+		return solve([
+                	x**3 + z**2 + y == v1,
+                  y**3 + x**2 + z == v2,
+                  z**3 + y**2 + x == v3,
+                  x + y + z == v4
+                ], x, y, z, solution_dict=True)[0]
+```
+
+We also choose to retrieve the solutions in a dictionary format, to make the solution parsing more convenient.
+
+Having the solutions, we can reconstruct the flag as:
+
+```python
+from Crypto.Util.number import long_to_bytes
+
+def get_flag(sols):
+		return b''.join([long_to_bytes(int(n)) for n in [sols[x], sols[y], sols[z]]])
+```
+
+## Getting the flag
+
+Let us recap the steps for getting the flag.
+
+- We noticed the flag is splitted into three parts.
+- We noticed that we are given four relations of these parts so the system must have a unique solution.
+- By minimal research, we see that we can solve systems using SageMath, SymPy or any online tool.
+- Solving for the three unknown parts, we reconstruct the flag.
+
+This recap can be represented with code as below:
+
+```python
+exec(open('output.txt').read())
+
+def pwn():
+    x, y, z = create_variables()
+    sols = solve_system(x, y, z)
+    flag = get_flag(sols)
+    print(flag)
+
+
+if __name__ == '__main__':
+		pwn()
+```
+

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/challenge/flag.txt

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+HTB{__protecting_the_secret_in_equations_is_not_secure__}

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/challenge/output.txt

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+v1 = 4196604293528562019178729176959696479940189487937638820300425092623669070870963842968690664766177268414970591786532318240478088400508536
+v2 = 11553755018372917030893247277947844502733193007054515695939193023629350385471097895533448484666684220755712537476486600303519342608532236
+v3 = 14943875659428467087081841480998474044007665197104764079769879270204055794811591927815227928936527971132575961879124968229204795457570030
+v4 = 6336816260107995932250378492551290960420748628

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/challenge/source.py

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+from Crypto.Util.number import bytes_to_long
+
+FLAG = open("flag.txt", "rb").read()
+
+step = len(FLAG) // 3
+candies = [bytes_to_long(FLAG[i:i+step]) for i in range(0, len(FLAG), step)]
+
+cnd1, cnd2, cnd3 = candies
+
+with open('output.txt', 'w') as f:
+    f.write(f'v1 = {cnd1**3 + cnd3**2 + cnd2}\n')
+    f.write(f'v2 = {cnd2**3 + cnd1**2 + cnd3}\n')
+    f.write(f'v3 = {cnd3**3 + cnd2**2 + cnd1}\n')
+    f.write(f'v4 = {cnd1 + cnd2 + cnd3}\n')

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/htb/output.txt

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+v1 = 4196604293528562019178729176959696479940189487937638820300425092623669070870963842968690664766177268414970591786532318240478088400508536
+v2 = 11553755018372917030893247277947844502733193007054515695939193023629350385471097895533448484666684220755712537476486600303519342608532236
+v3 = 14943875659428467087081841480998474044007665197104764079769879270204055794811591927815227928936527971132575961879124968229204795457570030
+v4 = 6336816260107995932250378492551290960420748628

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/htb/solver.py

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+from sage.all import *
+from Crypto.Util.number import long_to_bytes
+
+exec(open('output.txt').read())
+
+def create_variables():
+    x,y,z = var('x,y,z', domain=ZZ)
+    return x,y,z
+
+def solve_system(x, y, z):
+    return solve([
+        x**3 + z**2 + y == v1,
+        y**3 + x**2 + z == v2,
+        z**3 + y**2 + x == v3,
+        x + y + z == v4
+    ], x, y, z, solution_dict=True)[0]
+
+def get_flag(sols):
+    return b''.join([long_to_bytes(int(n)) for n in [sols[x], sols[y], sols[z]]])
+
+def pwn():
+    x, y, z = create_variables()
+    sols = solve_system(x, y, z)
+    flag = get_flag(sols)
+    print(flag)
+
+
+if __name__ == '__main__':
+    pwn()

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/release/crypto_sugar_free_candies/output.txt

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+v1 = 4196604293528562019178729176959696479940189487937638820300425092623669070870963842968690664766177268414970591786532318240478088400508536
+v2 = 11553755018372917030893247277947844502733193007054515695939193023629350385471097895533448484666684220755712537476486600303519342608532236
+v3 = 14943875659428467087081841480998474044007665197104764079769879270204055794811591927815227928936527971132575961879124968229204795457570030
+v4 = 6336816260107995932250378492551290960420748628

htb/hacktheboo2024/crypto/[Very Easy] sugar free candies/release/crypto_sugar_free_candies/source.py

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+from Crypto.Util.number import bytes_to_long
+
+FLAG = open("flag.txt", "rb").read()
+
+step = len(FLAG) // 3
+candies = [bytes_to_long(FLAG[i:i+step]) for i in range(0, len(FLAG), step)]
+
+cnd1, cnd2, cnd3 = candies
+
+with open('output.txt', 'w') as f:
+    f.write(f'v1 = {cnd1**3 + cnd3**2 + cnd2}\n')
+    f.write(f'v2 = {cnd2**3 + cnd1**2 + cnd3}\n')
+    f.write(f'v3 = {cnd3**3 + cnd2**2 + cnd1}\n')
+    f.write(f'v4 = {cnd1 + cnd2 + cnd3}\n')