- #1
christinamora
- 2
- 0
The question states prove,
If p is prime and p | a^n then p^n | a^n
I am pretty sure I have i just may need someone to help clean it up.
There are two relevant theorems i have for this.
the first says p is prime if and if p has the property that if p | ab then p | a or p | b
the second one is that if p is prime and p | a1a2a3...an, then p must divide one of the a_i.
so for the proof i am assuming p | a^n which i can rewrite as
p | a*a*a...an-1*an. so this is saying p(q) = a*a*a...an-1*an for some integer q.
now if I look at p^n | a^n that's the same as
p*p*p...pn-1*pn | a*a*a...an-1*an
well p(p*p*p...pn-1*p) | a*a*a...an-1*an
is that the way to go?
Or maybe before when i had that p(q) = a*a*a...an-1*an for some integer q.
just set q = p^n-1 so that p(q) = p^n.
I feel like the later way should do it.
Is this right?
If p is prime and p | a^n then p^n | a^n
I am pretty sure I have i just may need someone to help clean it up.
There are two relevant theorems i have for this.
the first says p is prime if and if p has the property that if p | ab then p | a or p | b
the second one is that if p is prime and p | a1a2a3...an, then p must divide one of the a_i.
so for the proof i am assuming p | a^n which i can rewrite as
p | a*a*a...an-1*an. so this is saying p(q) = a*a*a...an-1*an for some integer q.
now if I look at p^n | a^n that's the same as
p*p*p...pn-1*pn | a*a*a...an-1*an
well p(p*p*p...pn-1*p) | a*a*a...an-1*an
is that the way to go?
Or maybe before when i had that p(q) = a*a*a...an-1*an for some integer q.
just set q = p^n-1 so that p(q) = p^n.
I feel like the later way should do it.
Is this right?