Special number.

[Lan Todak]

Hi everyone... I have an interesting test to do. If given to you a^p, where p and a are available from 3 to infinity( although this set of numbers actually started from 0 but I removed them for some reasons), what is the special number(x) when you subtract from a, some composite numbers of a's reminder are always known? 

Here's an example. 
Pick any number for a,X and p. For this test I choose 8 for a, 11 for p and 31 for X so the structure looks like this, (8^11)-31=8,589,934,561. Can you guess the composite numbers for 8,589,934,561 without factorization? Some of its composite numbers are 13 and 660,764,197.
By using X =31, test X for every available number for a^p. Guess its composite numbers directly without using factorization.
If you are able to that, that means X is the special number. If u can't then it isn't.
Here are tips for you.
1. X value is always near one of the composite numbers ( that's why you can always guess them right away) within no more than 2 digits
For example, if X is 321, one of the composite numbers is within 300 to 350 width. More than that, you might calculate it wrong or X isn't special.
2. It works for any number for a^p. Yes, any number.
3. Use small numbers for test like 3^3-7 or 5^4-16 if you have issues with huge numbers

4. Remember to keep x as constant.only vary a  and p.

Good luck.

[Sensei]

14 minutes ago, Lan Todak said:

Guess its composite numbers directly without using factorization.

Rainbow table does it.

 

17 minutes ago, Lan Todak said:

2. It works for any number for a^p. Yes, any number.

Then write a computer program to show this.

 

[KJW]

Is this connected to Fermat's little theorem?

 

[Lan Todak]

8 hours ago, Sensei said:

Rainbow table does it.

 

Then write a computer program to show this.

 

I think you didn't get it. Should I say "any kind of machine calculation?". As I mentioned before, by looking at it, you could probably get the answer right away, with no machine involved (for example calculator).

Test it first. if nobody gets it, I will show what I've got.

8 hours ago, KJW said:

Is this connected to Fermat's little theorem?

 

I could say yes, but the X value is constant. It doesn't vary with a or p. Can you guess one of the composite numbers for 23^37-x. If X is 23, the composite numbers is 37, right? But you can't use 23 to other numbers. If a = X then yes.

[Lan Todak]

I think I should post this in puzzle section instead of here. The answer is 1. You can get this answer by rearranging Fermat's little theorem. 

(I think my question wasn't interesting enough because it's easy)