Here’s a more readable version

[https://drive.google.com/file/d/1mlEdKwohNG08ppDjZq1CpiUIag-6t03b/view?usp=sharing](https://drive.google.com/file/d/1mlEdKwohNG08ppDjZq1CpiUIag-6t03b/view?usp=sharing) About the existence of reverse mod function for exponential functions (problem with discrete logarithm). f (x) = a ^ (floor (x)) mod p This is a brief description of my research on these functions and a method by which we can find a function g (x) such that g (f (x)) = x The existence of such a function would make solving these kinds of problems trivial if a) g (x) exists and finding g (x) is possible b) calculating g (x) is the same as calculating f (x) (i.e. not computationally impossible) My research suggests that both a and b exist. That is, I found ag (x) to be trivial to calculate. However, I could only find such a g (x) for one very small prime number 5. That is as follows f (x) = 3 ^ floor (x) mod 5 g (x) = 〖4〗 ^ (floor ( x)) 〖 * 3〗 ^ (3 + floor (x)) mod 5 Interestingly, I’ve found g (x) for pretty much every f (x) I’ve viewed where the subgroup generated by a is p- 1. Where the subgroup generated by a is (p-1) / 2, we can also find g (x) by applying the same method, but for f ‘(x) = f (x) mod (p-1) / 2. However, this method of finding g (x) is neither easy to find nor easy to calculate, but it is very reliable to find g (x). The method to find g (x) is as follows We start with f (x) as follows f _1 (x) = a ^ (floor (x)) mod p Then we calculate F2 as f _2 (x) = a ^ (f _1 (x)) mod p Then we calculate F3 as f _3 (x) = a ^ (f _2 (x)) mod p We then calculate Fn as f _n (x) = a ^ (f _ (n-1) (x)) mod p We stop where Fn (x) = x. Now g (x) is equal to Fn-1, which means that g (f (x)) = x is being calculated. This means that solving the discrete logarithm problem can be done efficiently where we can find g (x) in the form given for the case a = 3 and p = 5. I have not yet been able to identify a mechanism to accelerate like me have found for the case of f (x) = 〖floor (x)〗 ^ e mod n Because in this case we can achieve the same behavior as described above we find g (x) de by applying f _n (x) = 〖f _ (n-1) (x)〗 ^ e mod n An interesting example of this is f (x) = 〖Floor (x)〗 ^ 5 mod 91 Where f (x) on its own g (x) is such that f (f (x)) = x, such that in terms of Fn n-1 would be equal to 1. In this case we can find Fn by f _n (x) = 〖f _1 (x)〗 ^ (e ^ n) mod n to calculate So for example in the case of f (x) = 〖floor (x)〗 ^ 5 mod 23 We get our g (x) is g (x) = 〖Floor (x)〗 ^ (5 ^ 4) mod 23 More interesting here is that there is a power n that g (x) works for all odd a for a given p. For example n = 9 is the universal (observed to be the lcm for all n of different a) n for 23. That is an odd a would have ag (x) as follows g (x) = 〖floor (x) 〗 ^ (A ^ 9) mod 23 Here is a table of some universal n + 1 that I calculated for numbers up to 50 (I have more but limited it here for readability) pnpnpnpn 5 2 17 4 29 6 41 2 6 1 18 1 30 2 42 2 7 2 19 6 31 4 43 6 8 1 20 2 32 2 44 4 9 1 21 2 33 4 45 1 10 2 22 4 34 4 46 10 11 4 23 10 35 2 47 22 12 1 24 1 36 1 48 2 13 2 25 4 37 6 49 6 14 2 26 2 38 6 50 4 15 2 27 1 39 2 16 2 28 2 40 2 This is a limited overview of some of the work I in the interesting field of mathematics, I have yet to identify ag (x) for the first problem which is in a similar format to that of the second problem shown. However, I believe there must exist ag (x) which is trivial to calculate (or at least the same as f (x)) my current work hint at the form of g (x) = (dynamic) ^ floor (x) mod p but I have yet to find something like this. I hope you enjoyed my brief overview of my independent research on this issue and look forward to your feedback. Thank you Zakrea Almansouri

[zakrea2070@gmail.com](mailto: zakrea2070@gmail.com)