Modular Arithmetic/Fermat's Little Theorem

Let us look at the powers of a number under modulo arithmetic. We'll look at:


 * $$\displaystyle 3, 3^2, 3^3, 3^4...$$

and we're going to look modulo a prime number $$\displaystyle p$$. First we'll choose $$\displaystyle p=7$$. We could work out each number and then reduce, e.g.


 * $$\displaystyle 3^4=3\times 3\times 3\times 3=81\equiv 4 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^5=3\times 3\times 3\times 3\times 3=243\equiv 5 \, \mathrm{mod} \, 7$$

but it is quicker just to work out $$\displaystyle 3^{n+1}$$ from $$\displaystyle 3^n$$ like so:


 * $$\displaystyle 3^1=3\quad \quad \quad \equiv 3 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^2=3 \times 3 =9 \equiv 2 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^3=2 \times 3 =6 \equiv 6 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^4=6 \times 3 =18 \equiv 4 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^5=4 \times 3 =12 \equiv 5 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^6=5 \times 3 =15 \equiv 1 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^7=1 \times 3 =3 \equiv 3 \, \mathrm{mod} \, 7$$
 * $$\displaystyle 3^8=3 \times 3 =9 \equiv 2 \, \mathrm{mod} \, 7$$

it is quite clear that the pattern is repeating. That's because $$\displaystyle 3^6\equiv1 \, \bmod \, 7$$. Even without doing the calculation it is clear that the pattern has to repeat somewhere. There are at most 7 different possible numbers for $$\displaystyle 3^n\,\bmod\,7$$ so if we calculate it for $$\displaystyle n=1,2,3,4,5,6,7,8$$ there has to be a repeated answer somewhere in there.

Note: This is a use of the pigeonhole principle - there are more pigeons than pigeonholes, so one pigeonhole must contain two pigeons. If you are interested, click on the pigeon icon to read more about the pigeonhole principle:



Once a number is repeated the sequence from there on must be the same as before, from the first occurrence of that number. We can even see that we will get a 1 followed by a 3 as where the repeat first happens. Why?

Suppose we have some repeated number, in other words $$\displaystyle 3^a\equiv3^{a+b}$$ with a and b positive numbers. Then $$\displaystyle 3^a\equiv3^a3^b$$ and we can divide both sides by $$\displaystyle 3^a$$ to get $$\displaystyle 1\equiv3^b$$, i.e. a repeat that happens sooner. Just a little care is needed. In modulo arithmetic it is not always allowed to divide by a common factor. We're allowed to do that division here because we earlier established it for modulo a prime using Euclid's algorithm. We know that $$\displaystyle 3^n\,\mathrm{mod}\,7$$ is not zero since otherwise 3 would be a factor of 7 and 7 is prime.

Something to watch out for with modular arithmetic, we cannot just reduce numbers wherever we see them. For example working $$\displaystyle \bmod \, 7$$, the exponent of $$\displaystyle 8$$ in $$\displaystyle 3^8$$ cannot just be replaced by $$\displaystyle 1$$ because $$ 3^8 \neq 3^1 \, \bmod 7$$. The ones to watch are in exponents. In expressions like $$\displaystyle 1000x \, \bmod 7$$ and $$\displaystyle (1000+x) \, \bmod 7$$ it is fine to replace the 1000 to get $$\displaystyle 6x \, \bmod 7$$ and $$\displaystyle (6+x) \, \bmod 7$$ or even $$\displaystyle -x \, \bmod 7$$ and $$\displaystyle (x-1) \, \bmod 7$$.


 * Now your turn. Do exactly the same thing as in the example, but this time for $$\displaystyle p=11$$

There is nothing special about 3 here. We could do exactly the same exercise for other numbers $$\displaystyle a$$ with $$\displaystyle a=1, 2, 4, 5 \, or \, 6$$. We might reach $$\displaystyle a^n\equiv1$$ sooner, we definitely will for $$\displaystyle a=1$$, but we would still have $$\displaystyle a^{(p-1)}\equiv1 \, \pmod p$$.

We will prove this several ways. The reason for making such a meal out of proving it, is that it helps to see different ways of proving a result. In this case, it's mainly a way to show the different notation that can be used. The third variant of the proof will also introduce the concept of multiplicative functions, which will be important later on.