# Question on the mod operator.

1. Feb 7, 2006

### srgut

If a + b = c, then is a + b = c (mod n) for all n?

For example while reading LeVeque's Topics in Number Theory I came across a section on Fermat's Last Theorem in which he says: a way to show c^n = a^n + b^n has no solution is to assume the infinite amount of congruences c^n = a^n + b^n (mod p) for p = 2, 3, 5, 7, ... and then derive a contradiction.

I assume what he means is: if c^n = a^n + b^n, then c^n = a^n + b^n (mod n) for any n that we choose.

Is this valid?

2. Feb 7, 2006

### Hurkyl

Staff Emeritus
What's the definition of equality, mod n?

It's ambiguous.

If you meant:

For any n: [ if c^n = a^n + b^n, then c^n = a^n + b^n (mod n) ]

That's fine. (Assuming you prove the theorem you originally asked)

If you meant

If c^n = a^n + b^n, then for any n: c^n = a^n + b^n (mod n)

the no, that's invalid. n already has a meaning, so you cannot introduce n as a dummy variable! You'd have to use a new letter, like p.

3. Feb 7, 2006

### srgut

Sorry that was clumsy of me, I didn't realize that I was using n as the variable for the exponent already. What I mean to say is:
If c^n = a^n + b^n, then under the mod equivalence c^n = a^n + b^n (mod k), where we can choose k at our leisure.

It might be trivial but it seems to me that something is missing between these two steps.

4. Feb 7, 2006

### Moo Of Doom

Yes, that's trivial. The definition of the mod k equivalence class is x = y (mod k) iff k|x-y. Thus c^n = a^n + b^n implies c^n-(a^n + b^n) = 0, which is divisible by k for any k (except 0).

5. Feb 7, 2006

### srgut

Ahh... of course that makes sense. I guess I just needed some reassurance to see it.

So then by convention zero is divisible by any n > 0, thats very interesting!

Thanks for the help.