Follow is a newly released RSS reader that is currently in beta testing and requires an invitation to gain access. Initially, invitation codes can only be obtained through official channels, such as the official team distributing a certain number of codes on Discord or X at irregular intervals. You can also directly email the official team or send private messages to obtain an invitation code.
At the same time, users who already have an invitation code and have activated it can invite other users, creating a viral spread of invitation codes.
- Check in daily to earn 20 powers; accumulate 100 powers to mint an invitation code.
- You can obtain one invitation code generation qualification every 3 days.
- Authenticating a subscription source grants you 100 powers directly, meaning a new user can generate a new invitation code in as little as 3 days.
New invitation codes generated by users generally have the following destinations:
- Invite friends and family
- Social media likes, comments, and giveaways, inviting lucky netizens (available on X, Xiaohongshu, Bilibili, etc.)
- As the answer to a riddle, "Matchmaking"
- Xianyu (a second-hand trading platform)
If you don't know anyone who can generate an invitation code for you and don't have enough luck to win one in a lottery among hundreds or thousands of people, you can find invitation codes on Discord, where many users post their generated codes. Since directly posting invitation codes can lead to some unscrupulous users snatching them up with scripts, people usually modify the invitation codes in some way, such as presenting them as images for speed typing or hiding one or two characters for luck. Such invitation codes typically last no more than five minutes.
To add an element of fun, people have started encrypting invitation codes in special ways, reminiscent of matchmaking. Some encrypt invitation codes using special encoding, some require solving puzzles to obtain the codes, some use graphic and text encryption, and others hide the codes within a story. During this time, I have been observing various decryptions that appear daily, and gradually I have been able to solve some of the riddles. However, since there are always people who solve them faster, I have never managed to obtain an invitation code through decryption, ultimately relying on sheer luck to guess one correctly. I have recorded some interesting decryptions as a memento of this time.
Graphic and Text Encryption#
This encryption method appears very abstract at first glance and requires a good understanding of commonly used graphic and text encryption content as well as imaginative thinking.
Pixel Art, Braille, Vigenere#
From @cap_stu
Hint 1: Using Photoshop is helpful
Hint 2: SOS
Hint 3: Two image ratios 1:800
Hint 4: One image with the word follow written on it
When this riddle was released, only the first three hints were given, which were too abstract to provide any clues. By the next day, no one had solved it, so the author provided hint four, at which point I began to notice some clues.
- First, look at hint four; below is the icon for the follow software, and above are six blocks of content. Associating this with Braille, I checked the table and found that the six blocks of content represent follow, confirming that the riddle uses Braille encryption.
- Decoding the stars above the riddle image using Braille yields
VIGENEREandPIXEL, indicating that another encryption method, Vigenere, is also needed, with the key being pixel. - The two gray lines above and below the icon at the bottom of the riddle can be decoded using Morse code, yielding Braille and Vigenere, which should have been hidden hints by the author before hint four was provided.
- The two segments of gray ciphertext in the middle of the riddle, when decoded using Morse code, yield
DRNAQXSDDQA. At this point, Vigenere comes into play, allowing for decryption to obtain the final answer.
| Braille Encryption: Replace to decrypt | Vigenere Encryption: Use key pixel to offset decrypt |
|---|---|
Attached is a diagram made by the author.
3D Graphics#
From @Plana
Riddle:
37.06749647,44.28503693,-40.50114678
39.18881682,45.3456971,-38.66402947
42.72435072,47.11346406,-35.60216729
63.93755416,57.72006577,-17.23099422
81.61522369,66.55890054,-1.921683328
...
The author provided a lot of such data, which is a CSV file. Since the data is grouped in threes, it was speculated to be RGB, but the author stated it is not. Therefore, it was inferred to be XYZ coordinates. Using Python to plot the results yields:
Genshin Impact Encryption#
The Teyvat language from Genshin Impact.
The cipher book is as follows, and the answer is sf4lBGG_7E.
Dancing People, Color Blocks, Tektite Language#
From @喵～
Not difficult; knowing the cipher book is enough to decrypt. The problem is that the author provided the wrong image, not distinguishing between uppercase and lowercase, leading to a direct decryption of EXSTCTTJKU. Fortunately, I guessed it correctly as eXsTCtTkKu and successfully activated it.
https://www.dcode.fr/ballet-alphabet
https://www.dcode.fr/hexahue-cipher
https://www.dcode.fr/tenctonese-alphabet
Moon, Base Conversion#
From @Jacob
- Hint 1: Moon + text + red = full code (implying that this code is a combination of three segments)
- Hint 2: Look closely; the two moons are different (the moon is the Lunar Alphabet, deciphering yields the first two letters)
- Hint 3: The numbers resemble "eight-part essay" (the numbers in the image suggest octal conversion, deciphering yields the second segment)
- Hint 4: The red lines conclude (the only line appearing in the code is the underscore _ for the third segment)
Solving the riddle:
- The moon should be solved using lunar phases, yielding
VD
- The image contains four lines of numbers, which are relatively large; since the numbers are in octal. Converting the octal numbers to decimal and then using Unicode encoding yields the answer: 【The two moon deities guide you to the dark side of the moon 5xGivZb red object】, revealing that the middle seven characters are
5xGivZb - The remaining red line is naturally the underscore
_
Thus, the invitation code obtained is: VD5xGivZb_
Conan Decryption#
From @Arona
AESDecrypt(data="vEIlU835MIeD08frTJmJpw==", "ECB", "PKCS7", "128bit", key=guess from the image)
The original question does not include the numbers on the left and the letters below; decrypting yields the answer EGG HEAD, which was originally the key, but the author copied it incorrectly during encryption, making it impossible to derive the invitation code.
Birds on Wires, Vigenere Encryption#
From @Jacob
Double encryption, the first layer being the birds on wires, with the cipher book being the image below (search for The bird on the telephone pole decrypt or similar content to find https://www.dcode.fr/birds-on-a-wire-cipher), where the position of the birds represents the case of the letters.
After decryption, it yields bHPwBTSiji, and then using Vigenere decryption with the key being the software name represented by the image, follow, yields the answer wTElNXNuyx.
Encoding Encryption#
This encryption typically requires knowledge of the ciphertext, key, and encryption method to decrypt.
Zero-width-web, SVG, QR Code#
From @Huajin, yes, I created this
Ciphertext:
M
- Hint 1: Double encryption
- Hint 2: f12, https://yuanfux.github.io/zero-width-web
- Hint 3: SVG
If you copy the ciphertext into WeChat or VSCode, you will find some characters that cannot be displayed. This is because this ciphertext hides zero-width characters. Using the tool provided in hint 2 can complete the first layer of decryption to obtain the real ciphertext.
M4 4.5h7m1 0h2m4 0h7M4 5.5h1m5 0h1m1 0h3m3 0h1m5 0h1M4 6.5h1m1 0h3m1 0h1m1 0h3m3 0h1m1 0h3m1 0h1M4 7.5h1m1 0h3m1 0h1m3 0h2m2 0h1m1 0h3m1 0h1M4 8.5h1m1 0h3m1 0h1m1 0h1m1 0h1m1 0h1m1 0h1m1 0h3m1 0h1M4 9.5h1m5 0h1m3 0h3m1 0h1m5 0h1M4 10.5h7m1 0h1m1 0h1m1 0h1m1 0h7M13 11.5h4M4 12.5h1m2 0h6m1
- Hint 1: Two layers of encryption
- Hint 2: f12, https://yuanfux.github.io/zero-width-web
- Hint 3: SVG
If you copy the ciphertext into WeChat or VSCode, you will find some characters that cannot be displayed. This is because this ciphertext hides zero-width characters. Using the tool provided in hint 2 can complete the first layer of decryption to obtain the real ciphertext.
M4 4.5h7m1 0h2m4 0h7M4 5.5h1m5 0h1m1 0h3m3 0h1m5 0h1M4 6.5h1m1 0h3m1 0h1m1 0h3m3 0h1m1 0h3m1 0h1M4 7.5h1m1 0h3m1 0h1m3 0h2m2 0h1m1 0h3m1 0h1M4 8.5h1m1 0h3m1 0h1m1 0h1m1 0h1m1 0h1m1 0h1m1 0h3m1 0h1M4 9.5h1m5 0h1m3 0h3m1 0h1m5 0h1M4 10.5h7m1 0h1m1 0h1m1 0h1m1 0h7M13 11.5h4M4 12.5h1m2 0h6m1
Combining these two segments of ciphertext reveals a large number of h and m. If you are familiar with SVG, you can immediately see that this is an SVG path tag. Simply search for an <svg>, pair it with the decrypted result <path d="..."/>, and you will obtain an SVG. Opening it reveals a QR code.
<svg viewBox="0 0 33 33" shape-rendering="crispEdges">
<path stroke="#fff"
d="M4 4.5h7m1 0h2m4 0h7M4 5.5h1m5 0h1m1 0h3m3 0h1m5 0h1M4 6.5h1m1 0h3m1 0h1m1 0h3m3 0h1m1 0h3m1 0h1M4 7.5h1m1 0h3m1 0h1m3 0h2m2 0h1m1 0h3m1 0h1M4 8.5h1m1 0h3m1 0h1m1 0h1m1 0h1m1 0h1m1 0h1m1 0h3m1 0h1M4 9.5h1m5 0h1m3 0h3m1 0h1m5 0h1M4 10.5h7m1 0h1m1 0h1m1 0h1m1 0h7M13 11.5h4M4 12.5h1m2 0h6m1
</svg>
You can also slightly beautify it by changing the background color to white and the color blocks to black.
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 29 29" shape-rendering="crispEdges">
<path fill="#ffffff" d="M0 0h29v29H0z" />
<path stroke="#000000"
d="M4 4.5h7m1 0h2m4 0h7M4 5.5h1m5 0h1m1 0h3m3 0h1m5 0h1M4 6.5h1m1 0h3m1 0h1m1 0h3m3 0h1m1 0h3m1 0h1M4 7.5h1m1 0h3m1 0h1m3 0h2m2 0h1m1 0h3m1 0h1M4 8.5h1m1 0h3m1 0h1m1 0h1m1 0h1m1 0h1m1 0h1m1 0h3m1 0h1M4 9.5h1m5 0h1m3 0h3m1 0h1m5 0h1M4 10.5h7m1 0h1m1 0h1m1 0h1m1 0h7M13 11.5h4M4 12.5h1m2 0h6m1
</svg>
Scanning the QR code, you will obtain the invitation code: VlbwY6fixN.
AES Encryption#
From @Oganneson
Ciphertext: YDY/4rmwped3BwBS8PYZjg==, AES 128-bit encryption, CBC mode, PKCS7 padding, with the key and initialization vector being the same.
This is not simple ciphertext; you can see many zero-width characters in the console.
Using https://yuanfux.github.io/zero-width-web to decrypt removes the zero-width characters and yields the real ciphertext YDY/4rmwped3BwBS8PYZjg== and the key bMfNvN74N8Kkd83p, after which you can use any AES decryption tool to obtain the invitation code.
Zen Buddhism Discussion#
From @Charles Ye
Riddle: The Buddha said: "Victory, nephew, punishment, overturn, white, fear, bowl, far, obtain, contempt, Mu, Ji, Buddha, monk, shine, nephew, contempt, punishment, Luo, dark, fear, nephew, luxury, one, nirvana, dream, fear, white, Jia, Xi, all, nephew, also, Brahman, participate, na, pan, no more."
Solution: It is obvious that this is encrypted with Zen Buddhism discussion; simply search for a riddle-solving website related to Zen Buddhism to decrypt.
New Buddha's Sayings Encryption#
From @dirt
New Buddha's Sayings: "All clerics fear falling, fear hum, all fear Tathagata, fear clerics, bowl, Sattva, fear wish, fear falling, mantra, pity, fear, truth, clerics, fear wisdom, mantra, fear, robbery, fear, Ah, is, honey, like."
Solution: A variant of Zen Buddhism discussion, you can decrypt it using this website http://hi.pcmoe.net/Buddha.html.
Caesar Encryption#
From @neoedon
Riddle: cRIrROPj5I has been Caesar encrypted, with an offset of 3, 5, 7, 9, 11, or 13; decrypt to obtain the invitation code.
Solution: The encryption method is known, simply search for a riddle-solving website that can decrypt Caesar encryption; it mainly depends on luck.
ASCII, Base Conversion, Offset#
From @cliouo
In an ancient castle, a mysterious letter is hidden. The letter begins with a string of encrypted characters:
65,171,142,147,161,40,153,164,161,161,164,142,152,151,40,147,144,40,154,70,162,161,124
The guardian of the castle tells you that this text uses a simple substitution encryption method, and you need to decrypt it to find the key to the treasure.
Hint:
The first character of the invitation code is a number N, and this text uses a simple substitution encryption method.
Each letter has been replaced with the letter that is N positions ahead of it in the alphabet.
Please decrypt this text and find the key to the treasure.
Solution: The numbers are octal ASCII codes; convert each value from octal to decimal, then look up its corresponding ASCII character. If it is a letter, offset by 5 (the first character decrypts to 5).
XTEA Encryption, App#
From @Deco
Ciphertext one: Zv2DGQAG2q+57dtXshJlzk7msqCY3M1LOU453/fLLboPUWiTUqmAs1iLbIvDkh7TvF1smYrpvZ4=
Ciphertext two (partial): Zv2DNQAJMOMfrw6ycIdOxF/meq9M5vf5
- Hint 1: XTEA encryption
- Hint 2: The key is the app name in lowercase
- Hint 3: The first riddle's key hint: Drank way too much coffee
- Hint 4: The first riddle's solution yields the second riddle's key hint
Solution: The encryption method is known; you just need to guess the key. The first key hint's content is youtube, which is the update announcement content for the app. The result of decryption is: "Fixed some bugs and ate a bucket of ice cream," which is the update announcement content for bilibili, serving as the latter half of the second riddle's ciphertext, making the key bilibili. This way, you can obtain the result.
Emoticon Encryption#
The ciphertext is very abstract, but it is not difficult; you can easily find a decryption tool.
ﾟωﾟﾉ= /｀ｍ´）ﾉ ~┻━┻ //´∇｀/ ['']; o=(ﾟｰﾟ) ==3; c=(ﾟΘﾟ) =(ﾟｰﾟ)-(ﾟｰﾟ); (ﾟДﾟ) =(ﾟΘﾟ)= (o^^o)/ (o^^o);(ﾟДﾟ)={ﾟΘﾟ: '' ,ﾟωﾟﾉ : ((ﾟωﾟﾉ==3) +'') [ﾟΘﾟ] ,ﾟｰﾟﾉ :(ﾟωﾟﾉ+ '')[o^^o -(ﾟΘﾟ)] ,ﾟДﾟﾉ:((ﾟｰﾟ==3) +'')[ﾟｰﾟ] }; (ﾟДﾟ) [ﾟΘﾟ] =((ﾟωﾟﾉ==3) +'') [c^^o];(ﾟДﾟ) ['c'] = ((ﾟДﾟ)+'') [ (ﾟｰﾟ)+(ﾟｰﾟ)-(ﾟΘﾟ) ];(ﾟДﾟ) ['o'] = ((ﾟДﾟ)+'') [ﾟΘﾟ];(ﾟoﾟ)=(ﾟДﾟ) ['c']+(ﾟДﾟ) ['o']+(ﾟωﾟﾉ +'')[ﾟΘﾟ]+ ((ﾟωﾟﾉ==3) +'') [ﾟｰﾟ] + ((ﾟДﾟ) +'') [(ﾟｰﾟ)+(ﾟｰﾟ)]+ ((ﾟｰﾟ==3) +'') [ﾟΘﾟ]+((ﾟｰﾟ==3) +'') [(ﾟｰﾟ) - (ﾟΘﾟ)]+(ﾟДﾟ) ['c']+((ﾟДﾟ)+'') [(ﾟｰﾟ)+(ﾟｰﾟ)]+ (ﾟДﾟ) ['o']+((ﾟｰﾟ==3) +'') [ﾟΘﾟ];(ﾟДﾟ) [''] =(o^^o) [ﾟoﾟ] [ﾟoﾟ];(ﾟεﾟ)=((ﾟｰﾟ==3) +'') [ﾟΘﾟ]+ (ﾟДﾟ) .ﾟДﾟﾉ+((ﾟДﾟ)+'') [(ﾟｰﾟ) + (ﾟｰﾟ)]+((ﾟｰﾟ==3) +'') [o^^o -ﾟΘﾟ]+((ﾟｰﾟ==3) +'') [ﾟΘﾟ]+ (ﾟωﾟﾉ +'') [ﾟΘﾟ]; (ﾟｰﾟ)+=(ﾟΘﾟ); (ﾟДﾟ)[ﾟεﾟ]='\'; (ﾟДﾟ).ﾟΘﾟﾉ=(ﾟДﾟ+ ﾟｰﾟ)[o^^o -(ﾟΘﾟ)];(oﾟｰﾟo)=(ﾟωﾟﾉ +'')[c^^o];(ﾟДﾟ) [ﾟoﾟ]='"';(ﾟДﾟ) [''] ( (ﾟДﾟ) [''] (ﾟεﾟ+(ﾟДﾟ)[ﾟoﾟ]+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟｰﾟ)+ ((o^^o) +(o^^o))+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ ((ﾟｰﾟ) + (ﾟΘﾟ))+ (c^^o)+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟΘﾟ)+ ((o^^o) - (ﾟΘﾟ))+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (o^^o)+ ((o^^o) - (ﾟΘﾟ))+ (ﾟДﾟ)[ﾟεﾟ]+((o^^o) +(o^^o))+ ((ﾟｰﾟ) + (o^^o))+ (ﾟДﾟ)[ﾟεﾟ]+((o^^o) +(o^^o))+ ((ﾟｰﾟ) + (o^^o))+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟΘﾟ)+ ((o^^o) - (ﾟΘﾟ))+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟΘﾟ)+ (c^^o)+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟΘﾟ)+ ((ﾟｰﾟ) + (o^^o))+ (ﾟДﾟ)[ﾟεﾟ]+(ﾟΘﾟ)+ (ﾟΘﾟ)+ (c^^o)+ (ﾟДﾟ)[ﾟoﾟ]) (ﾟΘﾟ)) ('');
Beast Language Encryption#
From @QistChan
~呜嗷嗷嗷嗷呜呜呜啊嗷呜嗷呜呜啊呜啊嗷啊呜~呜嗷呜~嗷~呜呜嗷嗷~嗷嗷嗷呜呜呜啊嗷呜嗷呜呜啊呜嗷嗷啊嗷啊呜~呜嗷嗷~嗷~呜嗷呜嗷啊嗷嗷嗷呜呜呜呜~呜嗷呜呜啊呜嗷嗷啊嗷啊呜~呜嗷啊~嗷~呜嗷呜嗷~嗷嗷嗷呜呜呜呜啊呜嗷呜呜啊呜嗷嗷啊嗷啊呜~呜~嗷~呜呜嗷嗷~嗷嗷嗷呜呜呜啊啊呜嗷呜呜啊呜~啊啊嗷啊呜~呜~啊~嗷~呜嗷呜呜嗷啊
ps. Due to the presence of the symbol ~ in the ciphertext, when posted on Discord, it has many strikethroughs.
First, replace the strikethroughs with ~, which is to convert using Markdown syntax, then use https://roar.iiilab.com/ to decrypt. After decryption, convert the result to ASCII characters in hexadecimal.
Morse Code#
From @LOOL
6 uppercase, 1 2 7 9 lowercase
Decrypted: cuKNXEwSvW
Prime Number Encryption/Decryption#
From @Jacob
This is a basic decryption; successfully decrypting requires posting the answer in the DC chat, and the first person to answer correctly will receive an invitation code.
19 71 71 53 67 71 11 43 47 61 5 47 41 79 23 11 83 3 2 23 37 2 61 7 2 43 5 11 19 2 53 53 97 71 47 41 2 43 7 29 11 61 61 97 17 23 13
Hint: There is only one layer of encryption.
Observing the numbers, it is easy to see that they are all prime numbers. By converting the ciphertext to letters according to the order of prime numbers, we obtain:
| Letter | Prime |
|---|---|
| a | 2 |
| b | 3 |
| c | 5 |
| ... | ... |
| z | 101 |
httpstenorcomviewbailardancehappytomandjerrygif
The corresponding website seems to be https://tenor.com/view/bailar-dance-happy-tom-and-jerry-gif. However, opening it reveals a blank page on the tenor website; the real answer still needs to piece together a string of numbers from the author's profile 14081211. Therefore, the real answer is:
https://tenor.com/view/bailar-dance-happy-tom-and-jerry-gif-14081211
The corresponding gif is tom and jerry.
Riddle Solving Encryption#
Advanced Mathematics#
From @Jonty Li, ① and ② correspond to the respective sequence numbers of the English letters.
The result is 6 and 3, corresponding to j and c, with the answer being ymyjfSgmMc.
Poetry#
From @toxomo
The myriad families light up, the world celebrates together.
Two lines of poetry, rhyming, each character used only once.
Solution: The myriad families light up, the world celebrates together. UKPOMdkNKW
Name Encryption#
From @qvKEAyHnx6
A very creative encryption, using the code as a name, which neither allows scripts to snatch it up instantly nor makes the decryption difficulty too high.
Game Encryption#
From @alphardex
This is even more heavyweight, directly creating a game https://mygo-find-app-code-2.netlify.app/
"Actually, my overall process is like this: play the game — use SL method to pass (or get lucky and pass directly) — get hints — appreciate CG for another hint — OCR (you can also search yourself) — Google Translate (or search directly) — decrypt."
Personal Website/Blog Decryption#
Many experts hide invitation codes in their blog articles and leave a line at the end saying, "There are .. invitation codes hidden in the text; can you find them?"
ps. Follow will soon enter public testing; do not leave comments under the blog asking for invitation codes, as this may trouble the blogger.
Quiet Forest - From Innei#
Moment - Quiet Forest (innei.in)
I subscribed to the personal website of the great Innei using Follow, and while reading the articles, I saw the author mention in the comments, "There are three Follow invitation codes hidden in the text," so I opened the console to start decrypting.
Searching for follow led me to this segment of ciphertext:
\u003c!-- eYb5bKTV3K, ZmSPCbUcTo --\u003e\u003cdiv hidden\u003e-.-- ..--- -... .. --- --- .---- -..- -.. -..- --..-- .---- ....- ..... --..-- -....- ...--\u003c/div\u003e
Here, \u003c represents <, and \u003e represents > (Unicode encoding). Replacing the encoded text with <> yields:
<!-- eYb5bKTV3K, ZmSPCbUcTo --><div hidden>-.-- ..--- -... .. --- --- .---- -..- -.. -..- --..-- .---- ....- ..... --..-- -....- ...--</div>
- The first half
<!-- eYb5bKTV3K, ZmSPCbUcTo -->is an HTML comment containing two invitation codes. - The second half is Morse code, where
<div hidden>is a div with the hidden attribute, which will be hidden on the page. Decoding yieldsY2BIOO1XDX,145,-3, and the invitation codes are likely uppercase, so the numbers should represent the positions of lowercase letters, yieldingy2BioO1XDX.-3likely indicates an offset (Caesar cipher), thus the invitation code isb2ElrR1AGA.
Looking through previous articles, I found other Follow invitation codes hidden.
React i18n CSR Best Practices - Quiet Forest (innei.in)
(By the way, this article hides an invitation code for Follow; can you find it?)
Since the great Innei seems to like hiding invitation codes in the page, we directly copied the entire page's HTML and then used regex matching.
(?<=\s)[A-Za-z0-9]{10}(?=\s)
It can be seen that the hiding method also uses HTML comments <!-- 1rsMVNWDIt -->.
Discussing the Design Philosophy of Follow - Quiet Forest (innei.in)
(This article hides two Follow invitation codes; can you find them?)
The title has been updated to include 4 hidden invitation codes; come find me!
Knowing the pattern, this time I directly matched -->.
Interestingly, this method only found one, while the remaining invitation codes were elusive. Later, I found comments saying how to decrypt each invitation code, but following their hints yielded no results. It seems I arrived late, and the content related to invitation codes had already been deleted by the author.
- Searching for "If you can see this, you are not far from success. Follow" yielded no matches.
- Searching for ciphertext 1
Caesar=3: trfdNUvNTuyielded no matches. - The images in the scroll bar: The images in the scroll bar are no longer available.
- I did not see the
alert()pop-up.
Timochan's Blog#
https://www.timochan.cn/notes/44# 总结
When the article was published, I happened to be reading an article in Follow and noticed that the end mentioned there is an invitation code hidden in the text:
There is one Follow invitation code in the text? Do you know where it is?
https://www.timochan.cn/api/objects/file/8r806hgw1r2acpofp7.png
At the end of the article, a URL for an image is enclosed with strikethrough and filter: blur(8px). When I first opened it, it was a QR code, and scanning it redirected back to the blog. Therefore, I initially interpreted it as an automatically generated reminder to include the original link when reprinting, which led me to overlook such an obvious hiding method. Eventually, I searched the entire text, checked the console for hidden content, and found nothing. I finally returned to this QR code. When I scanned it with WeChat, I didn't see the destination address, so I searched for a QR code decoding tool, and indeed, it revealed a string of ciphertext:
https://www.timochan.cn?follow_key=VTJGc2RHVmtYMThvQkpBYmVJaEFxakptUWNBRmxHa0Jta0Jad0QvUHMrND0=
Now that the ciphertext is known, I directly asked GPT if it could identify the encryption method used.
GPT told us that this is Base64 encoded data, and after successfully decoding it, I verified that the result provided by GPT was correct. GPT indicated that the new ciphertext's prefix indicates that it was generated using the OpenSSL encryption standard (such as AES), so I asked it to continue decoding using AES, with the key being 8 digits, guessing it to be the author's name timochan, but the decryption failed. Later, the author hinted that the key is in the text, consisting of 8 digits. Observing the date at the top of the article, 20241006, which happens to be eight digits, I provided it to GPT for decryption.
The result obtained is M8g1GosNyw. Since I had already activated it, I was unsure if this was the correct answer.