### Theory NO. 1 : Number Theory

19Feb08

Prove that for every natural number $n \geq$ there exists $n$ different natural numbers such as sum of any two is divisable by their diffrence.

Proof:
Because number $a+b$ is divisable by $a - b$, we can write $a+b \equiv 0 \mod{a-b}$ and of course $a-b \equiv 0 \mod{a-b}$. Adding those two together:
$a+b + a - b = 2a \equiv a - b \equiv \ 0 \mod{a-b} \Leftrightarrow 2a \equiv 0 \mod{a-b}$

Induction:
1. Sequence 1,2 works
2. Assume that sequence $x_{1}, x_{2}, ..., x_{n}$ works, then the number $2x_{1}x_{2}...x_{n}$ is divisable by the difference of any two numbers from the sequence
3. Let $p=x_{1}x_{2}...x_{n}$, then a sequence, including $n+1$ numbers:
$p, p + x_{1}, p + x_{2}, ..., p + x_{n}$
‘Works’.