在「Recovering a full PEM Private Key when half of it is redacted」這邊看到的，起因是 _SaxX_ 幫客戶做滲透測試時找到客戶公開在網路上的 SSH key，然後他就碼掉一部分貼出來：
This morning, I began another #pentest for a client. After some google-fu dorking combine with the major search engines, I found the id_rsa key that gave me access to the server and a bunch of others 😬😬
another mistake under their radar... pic.twitter.com/K99UyjKTzW
— S. A. X. X. (@_SaxX_) March 24, 2021
原圖是這樣，接下來就開始被還原 XD
首先是 OCR 的過程，被稱為是整個還原過程最難的一部分 (哭爸啊)：
Ironically, this was the hardest part of the challenge. It took the longest time of all the steps and was the easiest to make errors in.
接下來就是解讀 PEM 檔的格式，可以藉此得到裡面的參數。
然後是套公式，窮舉運算裡面的值，可以看到迴圈 kp 只算了 3 到 65537，就推算出可能的 p：
e = 65537
q = 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
dp = 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
for kp in range(3, e):
p_mul = dp * e - 1
if p_mul % kp == 0:
p = (p_mul // kp) + 1
if isPrime(p):
print(f"Possible p: {p}")
後面就是跑驗證確認，就被打出來了...
