Is this proof alright?

[daon]

This is just a lemma I need for a larger proof that I am working on...

Need to prove:
Prop: Let p, x, j, k, l be integers. If p does not divide (x + kp) then p does not divide x.

Pf: Assume BWOC that p divides x.
Then x = pj for some j. Now,
[P does not divide (x+kp)] \(\displaystyle \L \Rightarrow\) [(x+kp) \(\displaystyle \neq\) pl for some l]
\(\displaystyle \L \Rightarrow\) [(pj)+kp \(\displaystyle \neq\) pl] \(\displaystyle \L \Rightarrow\) [p(j+k) \(\displaystyle \neq\) pl]
Let l = (j+k) then: (can I do this with l?)
[p(j+k) \(\displaystyle \neq\) p(j+k)] *Contradiction*


Also, can I use this to show that:
[p does not divide (a + 1) - p] \(\displaystyle \L \Rightarrow\) [p does not divide (a+1)]?

Thanks,
Daon

 

[pka]

Prop: Let p, x, j, k, l be integers. If p does not divide (x + kp) then p does not divide x.
Pf: Assume BWOC that p divides x.
Then x = pj for some j.

Correct to this point!
Now, (x+kp)=(jp+kp)=(j+k)p. *Contradiction*

 

[daon]

Correct to this point!
Now, (x+kp)=(jp+kp)=(j+k)p. *Contradiction*

Why is this a contradiction?

edit: I also just realized I could have done it much more easily by contrapositive...

 

[pka]

If (j+k)p=(x+kp) then by definition, p divides x+kp.

 

[daon]

If (j+k)p=(x+kp) then by definition, p divides x+kp.

Didn't think it would be so simple, thanks .