Okee, here is some food for thought.
(7x-2) mod 9=0 when x=9K-1
K is an integer
This means that all number that can be expressed as 7(9k-1)-2 can always be divided evenly by 9. This isn't really complex, but it is fun if you like prime numbers

So, when:
x=9k-1     (7x-2) mod 9=0     (7x+7) mod 9=0
x=9k-2     (7x-4) mod 9=0     (7x+5) mod 9=0
x=9k-3     (7x-6) mod 9=0     (7x+3) mod 9=0
x=9k-4     (7x-8) mod 9=0     (7x+1) mod 9=0
x=9k-5     (7x-1) mod 9=0     (7x+8) mod 9=0
x=9k-6     (7x-3) mod 9=0     (7x+6) mod 9=0
x=9k-7     (7x-5) mod 9=0     (7x+4) mod 9=0
x=9k-8     (7x-7) mod 9=0     (7x+2) mod 9=0

When I did this, I learned some things about how numbers work. If you wanna have fun, try using values other than 7 and 9. Something this simple can reveal quite a bit about numbers.

Modular arithmetic, it was thought to me and I came up with two independent results:
- you can use modular arithmetic to find divisible criteria like you did, here is a simple example:
let n be a integer number in decimal form
let d(i) be the sequence of its digits in decimal form, i. e., n=sum from i=0 to floor(log(n))+1 of d(i)*10^i or n can be expressed as d(0)+d(1)*10+d(2)*10^2+...+d(n)*10^n
then you have
3 divides n if
3 divides sum d(i)*10^i, that is equivalent to
sum d(i)*10^i = 0 mod 3 <=>
sum d(i)*1^i = 0 mod 3 <=>
sum d(i) = 0 mod 3
so,
3 divides n if the sum of digits of n is divisible by 3

Exercises: do the same for 2, 10, 7 or any number you like.

- I noted what happened to (x+1)^n mod n when n is prime
tips: Pascal triangle, divisibility

Note I don't know how to write mathematics in english conventions, I did the best to translate... Also writing in plain text does not help.
Sorry if you don't understand but I wrote this quickly.

That is okay, I do not know how to properly "write math" yet. But yeah, my goal was to use that to test for primes. I'll look into this when my brain is functioning again, but I think I understand

Incredible how some people come up results already known, independently on math after learning the start of things.
And you are on a possible right track, although it gets complicated and you need results like Little Fermat Theorem and Euler's Theorem, many prime testing methods use congruences ideas.

I am curious. Are you thinking of following Mathematics to your superior course and career, or keep just as a hobby and possibly work on something related? Remember you are not obligated to answer or reply fast.

I am a math major in college and I do plan to go as far as to get a doctorate. I am not 100% sure, but I am fairly certain I will choose a math oriented career such as being a professor.

*By the way, Galandros, I'm Thunderbolt from UTI

Good luck then.

I know because the projects you show off here match those of ThunderBolt in UTI.

I would love to be a teacher for other things, but I would prefer to be a professor for math.

(m^2-n^2, 2mn, m^2+n^2) will generate all the pythagorean triples. For instance, m=2 n=1 gives (3,4,5).

Also, the set of all numbers that are part of a pythagorean triple isn't terribly interesting; all numbers larger than 2 are part of a pythagorean triple (3,4,5); (5,12,13); (7,24,25); etc. for odd numbers, even numbers just double these triples.

Oh... misinterpreted, sorry.

Well, I suppose you could do individual tests on m^2-n^2=N, 2mn=N, m^2+n^2=N...

In particular, the first two are factorings of N.

I'm still wondering though... even given the new conditions, we should still expect to see a LOT more than just 2 numbers in the first 10,000.

For example, 60 should work with (36, 48, 60); (45, 60, 75); (60, 80, 100).