1
$\begingroup$

I keep coming across proofs that seem to use the following derivation, but I'm unsure where it comes from. What theorem shows that this is a correct step to take?

$ g^{x} = g^y$ mod $p$ $\iff$ $x = y$ mod $p - 1$. Especially the $p-1$ part confuses me, obviously. I don't see why that should not be mod $p$.

I have reason to believe it has something to do with the Euler totient, but that could be wrong.

EDIT: I should add: $g$ is a primitive root.

$\endgroup$
2
$\begingroup$

If $g$ is a primitive root modulo $p,ord_pg=p-1$

If $g^x\equiv g^y\pmod p\iff g^{x-y}\equiv1\pmod p\iff ord_pg\mid (x-y)$ (Proof below)

$\implies (p-1)\mid (x-y)$

This can be generalized to any composite $m>4$ having primitive root.So, $m$ must be of the form $p^e,2p^e$ where $p$ is an odd prime.

Now, $\phi(p^e)=p^{e-1}(p-1), \phi(2p^e)=phi(2)\phi(p^e)=p^{e-1}(p-1)$

In either cases if $g$ is a primitive root modulo $m,ord_pg=\phi(m)=p^{e-1}(p-1)$

If $g^x\equiv g^y\pmod m\iff g^{x-y}\equiv1\pmod m\iff ord_mg\mid (x-y)$ $\implies p^{e-1}(p-1)\mid (x-y)$

[

Proof :

Let $ord_ma=d$ and $a^n\equiv1\pmod m$ and $n=q\cdot d+r$ where $0\le r<d$

So, $a^{q\cdot d+r}\equiv1\pmod m\implies a^r\cdot (a^d)^q\equiv1\implies a^r\equiv1\pmod m$

But $d$ is the smallest positive integer such that $a^d\equiv1\pmod m$

$\implies r=0\implies n=q\cdot d$ i.e., $d\mid n$

]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.