Find all integers such that $2 < x < 2014$ and $2015|(x^2-x)$ Find all integers, $x$, such that $2 < x < 2014$ and $2015|(x^2-x)$.
I factored it and now I know that $x > 45$ and I have found one solution so far: $(156)(155)= (2015)(12)$. It's just that I don't think I'm approaching it the right way.
I'm new to this kind of stuff and I'm just doing this for fun so any help please?

You need a couple of pieces of information to complete this.
First of all, a property of prime numbers: If $p$ is a prime number such that $p ~|~ ab$, then $p~|~a$ or $p~|~b$. Also $2015=5 \cdot 13\cdot 31$.
Now, if $2015~|~N$, then $5~|~N$. Since $2015~|~x(x-1)$, $5~|~x$ or $5~|~(x-1)$; thus $x\equiv0\pmod5$ or $x\equiv1\pmod5$.
Continue with the other two factors: $13~|~x(x-1)$, so $x\equiv0\pmod{13}$ or $x\equiv1\pmod{13}$. There are now four possibilities for $x$:


*

*$x\equiv0\pmod5$ and $x\equiv0\pmod{13}$

*$x\equiv0\pmod5$ and $x\equiv1\pmod{13}$

*$x\equiv1\pmod5$ and $x\equiv0\pmod{13}$

*$x\equiv1\pmod5$ and $x\equiv1\pmod{13}$


When you continue with the 31 factor, you will get eight possibilities.
Here's the second piece of information you need: The Chinese Remainder Theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem
Now, use the Chinese Remainder Theorem to find $x\mod (5\cdot13\cdot31)$ for each of the 8 cases, and find the solutions which are in the interval $[2,2013]$. (There are six of them, one of which is 156.)
