# Thread: Mod proof with primes

1. ## Mod proof with primes

If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

I think that we will have to consider cases, that is,

p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

I really don't know what I need to show this....

If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

I think that we will have to consider cases, that is,

p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

I really don't know what I need to show this....

If $[p]=0,2,4\!\!\!\pmod 6$ then p is even, if...etc.

Tonio

If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

I think that we will have to consider cases, that is,

p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

I really don't know what I need to show this....

Hey, did you prove this one....