I have tried to teach myself number theory, but I encountered a branch of mathematics that has no order as compared to Algebra. Number theory starts with the simple fact that k(k+1) is even. In other words two adjacent numbers multiplied together is an even number 2n. After that you can supposedly prove everything. I found that difficult. I, however, found one problem in the chapter one problem set interesting. It was simply to prove that if the sum of the digits of a number are divisible by 9 then the number is divisible by 9. I knew this to be true from my early arithmetic years, but now I was given a challenge to prove it. I attacked this problem from many directions and here is my result for all bases of numbers.
From time to time I get questions regarding change of base. This page gives instructions to change from one base to another.
Another very useful process is Euclid's Algorithm. It is commonly referred to as GCD or Greatest Common Divisor.I have investigated Squares. In some of the number theory problems and Mersenne Prime explorations Squares come up often. Here is a page with my ramblings on Squares.
Because of my computer background, I worked in binary, octal, and hexadecimal. It is easy to change to a power of two base such as 4, 8, 16, etc. by regrouping the binary bits. The regrouping is then the digit in the new base. By using the knowledge of the proof above, I could easily check with only addition of binary bits if a number has a factor of 3,7,15,31, etc. These are the Mersenne numbers 2n-1.
Recently I have been exploring the Little Fermat Theorem, which is a special case of Euler's Theorem. I have found an interesting result with respect to the Fermat Numbers: 22n+1 (3, 5, 17, 257 ...)1. Putting the proof to work I discovered the visual proof that 2(2n+1)+1 always has a factor of 21+1. This is true for any base: that B(2n+1)+1 always has a factor of B1+1.
2. 2a(2n+1)+1 = (2a+1)(the sum as k=0 to 2n of[(-1)k(2ak)]).
The proof for this is to change the base to 2a. Then the number has the form B(2n+1)+1 where B = 2a. Applying #1 above, the factor is B1+1. substituting 2a for B we are left with a factor of (2a+1).
Since, I don't want to actually subtract any numbers an algorithm to accomplish the same thing is appropriate:
Bx - By, x > y = sum as k = 0 to x-y-1 of [(B-1)B(y+k)]
3. 2an-1 = (2a-1)(the sum as k=0 to (n-1) of[2ak]).
The proof for this is the visual proof that 2 to a composite power has a composite power (an) number of digits all equal to 1. 26-1 = 1111112 = 6310. It is easy to see that this number is divisible by a group of 2 digits [22-1] or a group of three digits [23-1].
The general case for any base: Ban-1 = (Ba-1)(the sum as k=0 to (n-1) of[Bak]).
Proof: Let b = B-1 then B6-1 = bbbbbbB. The general case is divisible by all factors of the composite power including 1. (For base 2 this is the trivial 2-1 or 1). Therefore B6-1 is divisible by (B-1), (B2-1), or (B3-1)! Note: After dividing by B-1 the resulting number is a repeating digit of 1 in that base.
Example: 1012+1 = 1,000,000,000,001 = (103+1)10,000
a=4,n=1
Therefore 1,000,000,000,001 = (104+1)(100-104+108)=(104+1)(100+[(9)104+(9)105+(9)106+(9)107])=(10,001)(99990001)
From #2 above.
Here are some interesting base 10 tricks for finding numbers divisible by 11, 111=(3)(37), 1111=( (11)(101)), etc. The trick is to regroup the digits into 2, 3, 4, etc. digits and add them up. When the sum results in B-1 or a factor of B-1 then the number is divisble by the factors of B-1.
Example: 755,783
Regrouping by four digit groups: 75 5783
Summing the groups: 5858 = (58)(101)
10,000-1 = 9999 = (3)(3)(11)(101)
We can get carried away here by converting 58 to base 30: 1 28.
Adding the digits we get 29 which equals B-1.