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

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- Feb 8th 2010, 11:45 AMArczi1984Nonnegative integers problem
Let $\displaystyle N=30030$, which is the product of the first six primes. How many nonnegative integers $\displaystyle x$ less than $\displaystyle N$ have the property that $\displaystyle N$ divides $\displaystyle x^3-1$?

- Feb 10th 2010, 06:45 AMBingk
I'm not sure how to do this in a more direct way ... here's what I got so far:

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

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

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

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

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

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