Hi all, I'm stumped on this.
Prove that for prime and integer such as , we have :
I believe I have to show that , but I just don't know how to start this.
I've never stumbled across such a twisted problem yet ... thanks all
EDIT : I put the wrong exercise with the wrong question ... corrected.
Yes, I understand better now, thanks Defunkt.
I had an idea to show it though, could this be correct :
We know that
This can be written :
Dividing both sides by thus yields :
And thus :
Because and is prime, so cannot divide .
I don't know if this is a correct proof (in the methods used, for instance).
Note that you did not use the fact that m is prime anywhere - meaning that if this proof were true, then, specifically, . But take, for instance, , then:
Why did this happen? because from the numbers we can produce a new multiple of 4, ie. . Can you see how m being prime solves this problem, and then how to incorporate this into your proof (which is perfectly fine otherwise)?
Try to understand the problem we have here at first - it is trivial that the only element from that m divides is m itself, however we want to show that there arent any two (or more) elements in that set that, when multiplied together, would be a multiple of m (as was the case with m=4).
As I said in the beginning, this is actually equivalent to showing that .
To prove the claim, first note that . Now, prove the following lemma:
Let k be a prime and , then: . You can show this by induction or in a straight-forward way (try to make it as rigorous as you can). After you prove this, your proof will be finished once you apply the lemma to the current situation.