# Can this number theory problem be solved purely algebraically?

1. Feb 11, 2013

### Wiz14

Let p be a prime number and 1 <= a < p be an integer.

Prove that a divides p + 1 if and only if there exist integers m and n such that
a/p = 1/m + 1/n

My solution: a|p+1 then there exists an integer m such that am = p+1
Dividing by mp
a/p = 1/m + 1/mp
So if I choose n = mp(which is always an integer) am I done?

2. Feb 12, 2013

### dkotschessaa

The answer to your title question is, yes. These things are usually solved algebraically. Just requires a bit of reasoning as you can see.

It looks like a good start but it's not that clear to me. It doesn't mean your proof is wrong, it just means it doesn't convince me yet. My questions are:

Have we even used the fact that p is prime?

Have we really shown that m and n are integers?

Have we shown what the proof is asking in the sense that this is true both ways? (iff and only if?). Typically this requires you to structure your proof as a string of "iff" or "if and only if" statements, or to prove it both ways (if P then Q, then if Q then P).

-Dave K

3. Feb 12, 2013

### HallsofIvy

Your proof is fine one way. That is, you have proved "if a divides p+1, then there exist integers, m and n, such that a/p= 1/m+ 1/n".

But this is an "if and only if" statement. You have NOT YET proved the other way "if there exist integers, m and n, such that a/p= 1/m+ 1/n, then a divides p+1". That might be where you need to use "p is prime".

4. Feb 12, 2013