🧑‍🤝‍🧑 Removed the duplicate authors

Authors are already specified in the metadatas
fabricated-world · Apr 7, 2024 · 1361a29 · 1361a29
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diff --git a/posts/FCSCFibonacci.md b/posts/FCSCFibonacci.md
@@ -56,16 +56,14 @@ MOV R0,R3
 STP
 ```
 
-which gives us the machine code : 
+which gives us the machine code :
 
 ```
 800600014dad80010000800200018006000180040000CF004a53002100324dad0645C701CFF900301400
 ```
 
-which gives us the flag : 
+which gives us the flag :
 
 ```
 FCSC{770ac04f9f113284eeee2da655eba34af09a12dba789c19020f5fd4eff1b1907}
 ```
-
-by cartoone
diff --git a/posts/KnightCTFHello.md b/posts/KnightCTFHello.md
@@ -2,7 +2,7 @@
 published: true
 title: KnightCTF - Hello
 description: Our solution for the "Hello" challenge of the KnightCTF 2023
-author: Cartoone
+author: Cartoone and Kalitsune
 date: '01/22/2023'
 icon: https://ctftime.org/media/cache/34/9f/349fe0a23f72ce7efeca1cf6e9eed649.png
 tags:
@@ -24,5 +24,3 @@ Decoding this string, which is obviously in base 64, gives us :
 We can see that this string is encrypted with a substitution algorithm, as the structure seems correct. We look for the beginning of the key (we're told in the statement that the flag is coded in **vigenère**) as we know that the flag begins with KCTF we obtain KNIG we deduce that the key was KNIGHT and we obtain the flag:
 
 ```KCTF{h1_th3n_wh0_ar3_y0u}```
-
-Solved by cartoone and maple