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Let $ x $ be coprime to an odd prime $p$. Then consider

$$ f_p(x) = \frac{x^{p-1} - 1}{p} $$

By Fermat's little we know this is always an integer.

In 1850 Eisenstein proved that

$$ f_p(a \cdot b) = f_p(a) + f_p(b) \mod p $$

Question 1 : How to prove this ?

Question 2 : How did Eisenstein do it and how many proofs are there ? Are all proofs similar ??

Perhaps better posted as a new question but

Question 3 :

These fermat quotients resemble logarithms so i wonder if and when they are isomorphic to the modular logaritm with respect to some primitive root. That seemed like a natural question to me.

To my surprise I did not find anything for free about this online ?

mick
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1 Answers1

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Well here is a proof, I have no idea if it's how Eisenstein did it though.

Look at the difference between the two terms, we want to show this difference is divisible by $p$.

$\frac{a^{p-1}b^{p-1}-1}{p} - \frac{a^{p-1}-1}{p}-\frac{b^{p-1}-1}{p} = \frac{a^{p-1}b^{p-1}-a^{p-1}-b^{p-1}+1}{p} =\frac{(a^{p-1}-1)(b^{p-1}-1)}{p}$

But both terms of the numerator are divisible by $p$, so even when we divide by $p$ once it is still divisible by $p$. (Alternatively, this difference is just $pf_p(a)f_p(b)$)

As for the discrete logarithm question, I don't think anything like that can be true. For one thing $f_p(x)$ modulo $p$ doesn't just depend on $x$ modulo p. For example $f_5(1) = 0$ and $f_5(6)=259$.

Nate
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  • Wow that is a short proof. Makes me wonder why it was only proved in 1850 ? Probably because it had applications that he discovered around that time. As for the primitive root , solving values t such that $f_p(t) = 1 mod p $ should get a primitive root i think? Notice $ f_p( x + n p^2) = f_p(x) mod p $. not sure how to solve that efficiently . But food for thought. Maybe a road to THE primitive root. – mick Aug 23 '19 at 11:25