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

Printable View

- May 14th 2012, 03:21 PMalexandrabel90congruence
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 PMHallsofIvyRe: 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, $\displaystyle x^3- x+ 1= 1$ mod 6. - May 15th 2012, 01:35 AMSylvia104Re: Congruence
A little observation should tell you that $\displaystyle x^3-x+1$ is always odd and therefore can never be divisible by $\displaystyle 6.$ (Wink)

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