Generalization of an interesting congruence?

Question

So the basic question is : "$\forall u, v\in \mathbb{Z}$, $3$ divides $uv(u^2-v^2)$".

To prove this statement we look at the values of $u, v, \pmod3$

We obtain : $\bar{u}=\bar{0},\bar{1},\overline{-1}$ and $\bar{v}=\bar{0},\bar{1},\overline{-1}$. Then it gives $\bar{u}^2=\bar{0},\bar{1}$ and $\bar{v}^2=\bar{0},\bar{1}$. So for each value : $\bar{u}\bar{v}(\bar{u}^2-\bar{v}^2)=\bar{0}$.

Now I was wondering if for all $p>3$ a prime number then $p$ divides $uv(u^{p-1}-v^{p-1})$ ? It seems true with F Little T.

But here's my non formal approach. We consider $\mathbb{Z}/p\mathbb{Z}=\{\bar{0},...,\overline{p-1}\}=\{\overline{-\frac{p+1}{2}},...,\overline{-1},\overline{0},\bar{1},...,\overline{\frac{p+1}{2}}\}$. So, $\bar{u}$ and $\bar{v}$ can take all that values. Then if we take the successive exponentiation at each step we will obtain two $\bar{1}$ until the power $^{p-1}$. By combination we always obtain $\bar{0}$. How to formalize that ?

What do you think of this approach ?

Thanks in advance !

Answer 1

Well, if $gcd(u,p)=gcd(v,p)=1$, then by Fermat's little theorem: $$u^{\phi(p)}=u^{p-1}\equiv1\pmod p$$ $$v^{\phi(p)}=v^{p-1}\equiv1\pmod p$$ Hence $p\mid u^{p-1}-v^{p-1}$ and so $p\mid uv(u^{p-1}-v^{p-1})$.

Also if $p\mid u$ or $p\mid v$, then $p\mid uv$.

Answer 2

Yes. Consider the three cases:

1. $u \equiv 0 \mod p$,
2. $v \equiv 0 \mod p$,
3. $u^{p-1} \equiv v^{p-1} \equiv 1 \mod p$

Answer 3

It is easily true. When $u$ and $v$ are not multiples of $p$ you have$(u^{p-1}-v^{p-1})\equiv 0\pmod p$ and if one of them is multiple of $p$ you have $uv\equiv 0\pmod p$ Thus in all ocassion one has $uv(u^{p-1}-v^{p-1})\equiv 0\pmod p$