Language : C/C++
Platform : Windows x32
OS Version : ( Windows 7,10)
Packer / Protector : None

Description :

Goal : Find correct key.

Enter key in textbox and click OK button then you see result . 

Screenshot :

crackme_release.exe

Very nice crackme Salin!

Noticed couple of things and gave it a try,its using a recrusive function that looked like Fbonacci but with a twist - when n equals 8 it returns the first character of my input XORed with 0x78 instead of the normal Fibonacci value.

Code computes fib(14) and compares it against specific values in a switch statement. I knew standard fib(14) is 377 which equals 0x179. Working backwards since fib(8) gets replaced with char[0] ^ 0x78 , i needed to figure out what value makes the sequance land on 377.

Turns out fib(8) needs to be 21 for this to work. So char[0] ^ 0x78 = 21 which means char[0] = 21 ^ 0x78 = 0x6D = 'm'

There's a loop that validates characters 3 through 9 using a table at byte_411BFC.

The validation formula :

fib(i+8) - fib(i+6) == byte_411BFC[i] ^ char[3+i]

Just rearranged it to solve for each character and u get : a, m, b, i, q, u, e.

Now characters 1 and 2 were trickier. Character 1 determines a value dl through another Fibonacci call and character 2 gets XORed with that. There's a secondary check involving fib(17) that needs to equal 1597.

After some trial and error found that char[1] = 'o' gives the right Fibonacci value, and char[2] = 'z' satisfies the XOR constraint.

And if we finally put it all together we get a valid serial for this amazing crackme which is :

mozambique

I ran it through checksum calculation and got 0x12D4.

Some code where it compares :

0040134B | 893D 20434100 | mov dword ptr ds:[414320],edi |

00401351 | 81FF D4120000 | cmp edi,12D4 |

On 12/12/2025 at 9:04 PM, iced said:

Very nice crackme Salin!

Noticed couple of things and gave it a try,its using a recrusive function that looked like Fbonacci but with a twist - when n equals 8 it returns the first character of my input XORed with 0x78 instead of the normal Fibonacci value.

Code computes fib(14) and compares it against specific values in a switch statement. I knew standard fib(14) is 377 which equals 0x179. Working backwards since fib(8) gets replaced with char[0] ^ 0x78 , i needed to figure out what value makes the sequance land on 377.

Turns out fib(8) needs to be 21 for this to work. So char[0] ^ 0x78 = 21 which means char[0] = 21 ^ 0x78 = 0x6D = 'm'

There's a loop that validates characters 3 through 9 using a table at byte_411BFC.

The validation formula :

fib(i+8) - fib(i+6) == byte_411BFC[i] ^ char[3+i]

Just rearranged it to solve for each character and u get : a, m, b, i, q, u, e.

Now characters 1 and 2 were trickier. Character 1 determines a value dl through another Fibonacci call and character 2 gets XORed with that. There's a secondary check involving fib(17) that needs to equal 1597.

After some trial and error found that char[1] = 'o' gives the right Fibonacci value, and char[2] = 'z' satisfies the XOR constraint.

And if we finally put it all together we get a valid serial for this amazing crackme which is :

mozambique

I ran it through checksum calculation and got 0x12D4.

Some code where it compares :

0040134B | 893D 20434100 | mov dword ptr ds:[414320],edi |

00401351 | 81FF D4120000 | cmp edi,12D4 |

Congratulations!

I never imagined my old crackme would be solved after five years.

And now, I've come up with a new solution. It's 2026. As you know, we can use AI.

I sent Claude my challenge with the message, "Please solve this."

about 30 minutes later, Claude perfectly solved my challenge.

Im not sure if you solved it manually or with the help of AI, but either way, I'm grateful.

2 hours ago, Salin said:

about 30 minutes later, Claude perfectly solved my challenge.

Please can you post the result?

Ted.