Thread: 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

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, mod 6.

A little observation should tell you that is always odd and therefore can never be divisible by

If you delve further, you have a product of three consecutive integers – this is always divisible by because it always contains exactly one multiple of and at least one multiple of Hence for all