Cool proof, but incomplete I need help.

Question

kainui · Answer

Ok, so I need help showing part of the proof of this. Does the following diophantine equation have nontrivial integer solutions mod p (a prime)? 

$$ax^2+by^2+cz^2 \equiv 0 \mod p$$
Note, trivial being we always have this solution $x=y=z=0$.

Let's find out, well first we make the AMAZING step and recognize by Fermat's little theorem that when x,y,z IS NOT a solution:

$$(ax^2+by^2+cz^2)^{p-1}  \equiv  1 \mod p$$

But when IT IS a solution we get what we'd expect:

$$(ax^2+by^2+cz^2)^{p-1} \equiv 0 \mod p$$

So now we can do this!

$$N \equiv \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \sum_{z=0}^{p-1} (ax^2+by^2+cz^2)^{p-1}  \mod p$$

N is the total number of Non solutions! Real fast let's just think about this, there are in total $p^3$ Possible solutions so the total number of solutions is:

$$p^3-N = \text{number of solutions}$$

Now here's where it gets a bit unclear to me, so I will state without proof that:

$$N \equiv 0 \mod p$$

Knowing this means that the number of solutions is divisible by $p$! We may be tempted to say there are 0 solutions, but we ALWAYS have the trivial solution, so there are at least $p-1$ other solutions!

kainui · Answer

So yeah, haha... If you don't think that's cool, GET OUT! xD The thing I need help showing (I sorta have some ideas but idk) is this:

$$ \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \sum_{z=0}^{p-1} (ax^2+by^2+cz^2)^{p-1} \equiv 0 \mod p$$
To make it nicer, I'll just write it as just to save space:

$$\sum_{x,y,z}(ax^2+by^2+cz^2)^{p-1} \equiv 0 \mod p$$

Anywho, there it is. If you want me to explain the proof more I'd be happy to help show you why it's cool.

ilovepuppieslol · Answer

bull, u dont need my help

kainui · Answer

@ganeshie8

ganeshie8 · Answer

Basically you're trying to show that $p$ divides the number of "non solutions" $N$.

$p\mid N$

If that were true, then the number of solutions $p^3-N = p^3-pu$ would also be divisible by $p$. Interesting...

ganeshie8 · Answer

Just went through your second reply... 
Looks you already have a proof for $p\mid N$  ?

kainui · Answer

Yeah exactly, this is not something I made up, it's something I found in one of the books I'm reading. I didn't really get their explanation of why that sum is divisible by p, so I thought I'd come here. I can copy paste the proof they give here though.

ganeshie8 · Answer

Ohk... both results look really cool.. I'll just give it a quick try, one moment..

kainui · Answer

Yeah, I will give it another try as well.

ganeshie8 · Answer

Here is what I have so far

Since $p$ is a prime, there exists a primitive root $r$ for $p$. This means we can express every integer in $\{1,2,3,\ldots,p-1\}$ as a power of $r$.

ganeshie8 · Answer

Looking at $(ax^2+by^2+cz^2)^{p-1}$,

Let 
$\large  a \equiv  r^{e_a} \pmod{p}$
$\large  b \equiv  r^{e_b} \pmod{p}$
$\large  c \equiv  r^{e_c} \pmod{p}$
$\large  x \equiv  r^{e_x} \pmod{p}$
$\large  y \equiv  r^{e_y} \pmod{p}$
$\large  z \equiv  r^{e_z} \pmod{p}$

ganeshie8 · Answer

$$\large  (ax^2+by^2+cz^2)^{p-1}\equiv  (r^{e_a+2e_x}+r^{e_b+2e_y}+r^{e_c+2e_z})^{p-1}\pmod{p}$$

kainui · Answer

$$\sum_{x,y,z} (ax^2+by^2+cz^2)^{p-1}$$
$$\sum_{x,y,z}\sum_{i, j,k} \frac{(p-1)!}{i!j!k!} a^ib^jc^k x^{2i}y^{2j}z^{2k}$$

The indices on i,j,k always satisfy the relation $i+j+k=p-1$ and every possible permutation of that is accounted for in the sum, which is not easy to write down as a simple sum from start to end. At the very least, we know that each of the $x^{2i}$ will take on the exponents from 2i = 0, 2, 4, ..., 2(p-1) at some point. However we can rearrange the sum and pick out x, y, or z of a given exponent and exchange the sum in this way:

$$\sum_{y,z}\sum_{i, j,k} \frac{(p-1)!}{i!j!k!} a^ib^jc^k y^{2j}z^{2k} \sum_{x=0}^{p-1} x^{2i}$$
For ease of reading, I'll replace all the i,j,k only dependent stuff with simply an $\alpha$.

$$\sum_{y,z}\sum_{i, j,k} \alpha_{ijk} y^{2j}z^{2k} \sum_{x=0}^{p-1} x^{2i}$$

Now the vague way in which I've said this (basically trying to relay what I've read) they end up saying because this is true:

Let $0 \le n < p-1$ then:

$$\sum_{x=0}^{p-1} x^n \equiv 0 \mod p $$

I should mention, this is not the geometric series. I will explain my reasoning for this being true (kinda iffy) in the next post.

kainui · Answer

Basically, you end up doing this for all the terms, and you get 0, which shows N=0 right? So onto my explanation.

So since multiplication in mod p we have a group, if we take every element and raise it to the same power we should end up with a permutation of the elements of the group but in a different order, so we can rewrite:

$$\sum_{x=0}^{p-1} x^n =\sum_{x=0}^{p-1} x \mod p$$

Now since addition in mod p is also a group, that means we have all the elements and every element has a unique inverse so to sum over them all means we will have summed over each element and its unique inverse (unless that element is its on inverse, but that won't happen since p is prime) so the sum is zero!

That's the proof I came up with, but it seems sorta sketchy mainly cause I'm more or less using ideas I'm just used to but I am not sure I can prove that mod p arithmetic is a field; I've never done it before at least.