Update 20240618_0.md

src/pages/blog/posts/20240618_0.md

@@ -29,7 +29,56 @@ thumbnail: "https://github.com/Liberluna/liberluna.github.io/tree/main/public/as
 | Tag | Beginner |
 | Desc | Using a safe prime makes RSA secure, doesn't it? |
 | 配布ファイル | Safe_Prime.tar.gz 774280c0d7d278ed01f537b13014fa15b4dc1d3a |
+ソースコード
+```python
+import os
+from Crypto.Util.number import getPrime, isPrime
 
-$n = p \cdot (2 p + 1)$ で構成されています。
+FLAG = os.getenv("FLAG", "ctf4b{*** REDACTED ***}").encode()
+m = int.from_bytes(FLAG, 'big')
+
+while True:
+    p = getPrime(512)
+    q = 2 * p + 1
+    if isPrime(q):
+        break
+
+n = p * q
+e = 65537
+c = pow(m, e, n)
+
+print(f"{n = }")
+print(f"{c = }")
+```
+$n = p \cdot (2 p + 1)$ で表されます。
 </br>
-$n$ が $p$ のみで決定するので，二分探索します・
+$n$ から $p$ を求めることができるのでsolverを書きます。
+</br>
+※PyCryptoDomeを使っても解けます。
+```python
+from sympy import nextprime, isprime
+from math import isqrt
+
+n = 292927367433510948901751902057717800692038691293351366163009654796102787183601223853665784238601655926920628800436003079044921928983307813012149143680956641439800408783429996002829316421340550469318295239640149707659994033143360850517185860496309968947622345912323183329662031340775767654881876683235701491291
+c = 40791470236110804733312817275921324892019927976655404478966109115157033048751614414177683787333122984170869148886461684367352872341935843163852393126653174874958667177632653833127408726094823976937236033974500273341920433616691535827765625224845089258529412235827313525710616060854484132337663369013424587861
+
+p = isqrt(n // 2)
+while True:
+    q = 2 * p + 1
+    if n % p == 0 and n % q == 0 and n == p * q:
+        break
+    p = nextprime(p)
+
+q = 2 * p + 1
+phi = (p - 1) * (q - 1)
+e = 65537
+d = pow(e, -1, phi)
+
+m = pow(c, d, n)
+flag = m.to_bytes((m.bit_length() + 7) // 8, 'big').decode()
+
+print("FLAG:", flag)
+```
+```
+ctf4b{R3l4ted_pr1m3s_4re_vuLner4ble_n0_maTt3r_h0W_l4rGe_p_1s}
+```