# congruence

• May 14th 2012, 03:21 PM
alexandrabel90
congruence
find all x such that x^3-x+1 is congruent to 0 mod 6.

i tried but it seems like there is no x
• May 14th 2012, 05:11 PM
HallsofIvy
Re: congruence
You are right.

0^3- 0+ 1= 1
1^3- 1+ 1= 1
2^3- 2+ 1= 7= 1 mod 6
3^3- 3+ 1= 25= 1 mod 6
4^3- 4+ 1= 61= 1 mod 6
5^3- 5+ 1= 121= 1 mod 6
6^3- 6+ 1= 211= 1 mod 6
That's all you have to do. For all x, $x^3- x+ 1= 1$ mod 6.
• May 15th 2012, 01:35 AM
Sylvia104
Re: Congruence
A little observation should tell you that $x^3-x+1$ is always odd and therefore can never be divisible by $6.$ (Wink)

If you delve further, you have $x^3-x=(x-1)x(x+1),$ a product of three consecutive integers – this is always divisible by $6$ because it always contains exactly one multiple of $3$ and at least one multiple of $2.$ Hence $x^3-x+1\equiv1\mod6$ for all $x\in\mathbb Z.$