According to Fermat's Little Theorem, for all integer $a$, if $p$ is a prime, then $$a^p \equiv a \pmod p$$ In other words, there exists a non-zero integer $k$ such that $$a^p-a=pk$$ Is there a method to determine $k$? I have seen many proofs using the multinomial expansion and/or recurrent analysis but none explicitly mentioned what $k$ is? Any Hints?
EXPLANATION:let me be clearer, let's say, for any 2 integers $a,b$, if $p$ is odd, then $$(a+b)^p\equiv {a^p+b^p}\pmod p$$ In other words, there exists $k \in \mathbb {Z} \neq 0$ such that $$(a+b)^p=a^p+b^p+kp$$ In this case using the binomial expansion, it is easy to see that $$k=\dfrac{\sum_{k=1}^{p-1}{ n\choose k}a^{p-k}b^k}{p}$$ I was wondering if there is a similar expression of $k$ in the Fermat's Little Theorem.
You are stating an equivalence between those two statements, namely $$a^p\,=\,a\;(mod\,p) \Longleftrightarrow\,a^p\,-\,a\,=p\,k$$ This means, that second statement is the definition of $k$.
PS:
1)Your example is not different as it is still giving $k$ in terms of the two sides and $p$. Thus it is not a method for calculating $k$ but it's merely re-expressing $k\,=\,\frac{(a+b)^b\,-\,(a^p+b^p)}{p}$ differently.
You could as well write $k=\frac{a\,(a^{p-1}-1)}{p}$ for the first relation. Not knowing how $a$ (and $b$ as well in your example) depends on $p$ the best you can do is give a formal expression for $k$ -as you do in your example.
2)If what you want is to still write it in terms of some power series you could argue as follows. Given that $\sum^{p-1}_{n=0}a^n\,=\,\frac{a^{p}-a}{a-1}$ for $a\neq1$, then $$k\,=\,\frac{(a-1)\,\sum^{p-1}_{n=0}a^n}{p}$$ But this doesn't provide you with a method for calculating $k$ different from directly using its definition.