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My question is specifically regarding congruence and the use of cases with the definition. I am only using the below mentioned example to emphasis my question.

I have a definition of congruence in my notes but I think I did not copy it down correctly.

Thm: $\forall a \in \mathbb{Z}, n \in \mathbb{N}, \exists q \in \mathbb{Z}, r \in \{0,1,...,(n-1)\}\,\textrm{such that}\,a= nq+r...$ I think I'm missing something in my definition.

I'm just a bit confused why do we have these cases in the proof? Can someone please comment on the use of cases?

if $a \in \mathbb{Z}$, then $a^3 \equiv a\ (\textrm{mod}\ 3)$.

\begin{align} a & \equiv r\ (\textrm{mod}\ n)\\\\ a^3 &\equiv 0\ (\textrm{mod}\ 3)\\ a^3 &\equiv 1\ (\textrm{mod}\ 3)\\ a^3 &\equiv 2\ (\textrm{mod}\ 3) \end{align}

Bill Dubuque
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Ziggy
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  • This is Fermat's little theorem that can easily be proven by induction. For such a small exponent however you can also consider $$a^3-a=a(a-1)(a+1)$$ three consecutive integers , if $a$ is an integer , so exactly one of them must be divisible by $3$. – Peter Sep 20 '23 at 15:20
  • I did expand this to get what you have, but I didn't consider that they must be divisible by 3. Why must they be divisible by 3? How can I see the consecutive integers? Why do we have three cases? – Ziggy Sep 20 '23 at 15:23
  • $a-1$ , $a$ and $a+1$ are the consecutive integers ($a$ is an integer). Consider the three possible cases of the residue class of $a-1$ modulo $3$ and find out which number is divisible by $3$ in each case. – Peter Sep 20 '23 at 15:25
  • Reading about congruence I have stumbled upon the word residue, but that has not been covered in any of my university courses yet. I can see your point now. But by definition of congruence $a$ can only equal ${0,1,2}$ – Ziggy Sep 20 '23 at 15:26
  • In general , you can say for every integer $n\ge 2$ : Exactly one out of $n$ consecutive integers must be disivible by $n$ because they form a full representation system modulo $n$. – Peter Sep 20 '23 at 15:27

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