Find the secret key for this function

[Syoma]

Originally Posted by Kerlingen

RSA is x^y with known x and unknown y.
This is x^y with unknown x and known y.
Two completely different things.

RSA powmod primitive is
c = m^e mod N, where c is ciphertext, m - message, e - public exponent, N - modulus
m = c^d mod N, where d is private exponent

our case is
0x90de757572b51cd3 = key^7 mod (max_uint64+1)

[UniSoft]

Originally Posted by Syoma

RSA powmod primitive is
c = m^e mod N, where c is ciphertext, m - message, e - public exponent, N - modulus
m = c^d mod N, where d is private exponent

our case is
0x90de757572b51cd3 = key^7 mod (max_uint64+1)

This is not the RSA !!!
In RSA N = p * q, where p and q are prime numbers!!!
max_uint64+1 = 0x10000000000000000 is not the product of two prime numbers, therefore the equation m = (m^e mod N)^d mod N will not work (mainly because of the large number of collisions).

[Syoma]

Originally Posted by UniSoft

This is not the RSA !!!

Did you ever read the topic? I wrote
It is the standard x^y mod n = c like RSA uses but with bad modulo N.
I did not write it is RSA. My suggestion was there may be simple math solution in number theory.

Just reviewed most popular directions. Do not see any simple solution.
x^7 = 3*3*31*4129*1499933*6041353 mod 2^64

[DARKER]

0x90de757572b51cd3 = 10438910133587483859

10419912285387762041 = 521 ^7
10438910133587483859 = ~521,136 ^7 <- this is your number
10560719825892021888 = 522 ^7

but your key is uint64, so this is not possible because ~521,136 is not integer

PS: Other solution can be some another integer, it's case when you strip result to uint64 (last bytes of result are 0x90de757572b51cd3 - it's just end of some big number)