# Nonnegative integers problem

• Feb 8th 2010, 11:45 AM
Arczi1984
Nonnegative 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 AM
Bingk
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\$