1. ## Nonnegative integers problem

Let $N=30030$, which is the product of the first six primes. How many nonnegative integers $x$ less than $N$ have the property that $N$ divides $x^3-1$?

2. I'm not sure how to do this in a more direct way ... here's what I got so far:

$x^3-1=(x-1)(x^2+x+1)$

So, $N|x-1$ or $N|x^2+x+1$

The first case gives us $x=1$, and nothing else since $x<30030$

The second case has no solutions, and I may be wrong here somewhere (it's late in my part of the world )

$N|x(x+1) + 1$, since N is the product of the first 6 primes, this includes 2 ... so $2|x(x+1) + 1$. But, $x(x+1)$ will be even, and by adding 1, you get an odd number, so 2 can't divide $x(x+1) + 1$.

So ... I believe the answer is that there is only 1 solution, which is when $x=1$