Given integers \(a\) and \(b\), and a natural number \(c\), we say that "\(a\) is congruent to \(b\) modulo \(c\)" and we write $$ \begin{align*} a \equiv b \bmod c \quad \text{or} \quad a \equiv_c b \end{align*} $$ if \(a - b\) is divisible by \(c\) or \(c \ | \ a - b\).

Addition and multiplication are preserved module a fixed number so

Let \(a_1, a_2, b_1, b_2\) be integers with \(a_1 \equiv a_2 \bmod c\) and \(b_1 \equiv b_2 \bmod c\). Then $$ \begin{align*} a_1 + b_1 &\equiv a_2 + b_2 \bmod c \\ a_1b_1 &\equiv a_2b_2 \bmod c \end{align*} $$

Note that division is not preserved. Similarly \(a_1^{b_1} \not\equiv a_2^{b_2} \bmod c\)

Residue Classes

Based on the definition of congruence, we can define residue classes as follows

For each integer \(a\), write $$ \begin{align*} [a] &= [a]_n = \{x \in \mathbb{Z} \ | \ x \equiv a \bmod n\} \subseteq \mathbb{Z} \\ &= \{ a + ny \ | \ y \in \mathbb{Z} \} \end{align*} $$ The set \([a]\) is called the residue class or congruence class of \(a\) modulo \(n\).

Example: for \(c=5\), the set of residue classes are \([0],[1],[2],[3],[4]\) and \([0] = [5] = [10] = \cdots\) since they all leave the same remainder modulo 5.

Both addition and multiplication are defined on congruence classes.

Define operations \(+\), \(\cdot\) on \(\mathbb{Z}_n\) by $$ \begin{align*} [a]_n + [b]_n &= [a + b]_n \\ [a]_n[b]_n &= [ab]_n \end{align*} $$

The set of residue classes form a ring.

Example 1

Is \(357\) divisible by 9?

We typically do this by summing the digits \(3+5+7 = 15\) and then again \(1+5 = 6\). From this we conclude that \(357 \equiv 6 \bmod 9\). Why does this work? because

$$ \begin{align*} 357 = 3 \cdot 10^2 + 5 \cdot 10^1 + 7 \cdot 10^0 \end{align*} $$

We know that \(10 \equiv 1 \bmod 9\). This means that

$$ \begin{align*} 10^k &\equiv 1^k \equiv 1 \bmod 9 \end{align*} $$

Thus

$$ \begin{align*} 357 &\equiv 3 \cdot 10^2 + 5 \cdot 10^1 + 7 \cdot 10^0 \bmod 9 \\ &\equiv 3 \cdot 1^2 + 5 \cdot 1^1 + 7 \cdot 1^0 \bmod 9 \\ &\equiv 15 \bmod 9 \\ &\equiv 6 \bmod 9 \end{align*} $$

Therefore, it is not divisible by 9.

Example 2

Is \(1234\) is divisible by 11?

We do this by taking the alternating sum of the digits so \(4 - 3 + 2 - 1 = 2\). Thus, \(1234 \equiv 2 \mod 11\) and \(1234\) is not divisible by 11. This works because

$$ \begin{align*} 1234 = 1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0 \end{align*} $$

But we know that

$$ \begin{align*} 10 \equiv -1 \bmod 11 \end{align*} $$

Therefore

$$ \begin{align*} 1234 &\equiv 1 \cdot (-1)^3 + 2 \cdot (-1)^2 + 3 \cdot (-1)^1 + 4 \cdot (-1)^0 \bmod 11 \\ &\equiv (1 \cdot -1) + (2 \cdot 1) + (3 \cdot -1) + (4 \cdot 1) \bmod 11 \\ &\equiv -1 + 2 - 3 + 4 \bmod 11 \end{align*} $$

Example 3

Is \(1234567\) a square?

This means that we want to know if \(1234567 = a^2\) for some \(a\). Then, suppose \(b\) is the last digit of \(a\). If we take \(a\) module \(10\), then we will get exactly \(b\). Thus

$$ \begin{align*} a &\equiv b \mod 10 \\ a^2 &\equiv b^2 \mod 10 \quad \text{(taking both sides to the same power)} \end{align*} $$

We know \(b\) is a single digit. Then \(b\) can only be some number in \(\{0,1,2,3,4,5,6,7,8,9\}\). Thus, \(b^2\) can only be one of the following

$$ \begin{align*} 0,1,4,16,25,36,49,64,81 \end{align*} $$

But we’re working module \(10\), so the possible unique digits are \(\{0,1,4,9,6,5\}\). Thus, 7 is not in the set so \(123457\) is not a square.

Example 4

What is \(a^2 \bmod 8\)?

If we take \(a\) module 8, then the possible distinct digits are \(0, 1,2,3,4,5,6,7\). So now if we take \(a^2\) module 8, we get

$$ \begin{align*} a^2 \equiv 0, 1, 4, 1, 0, 1, 4, 1 \bmod 8 \end{align*} $$

An an application of this, we can ask: can we write integers as sums of three squares so \(a^2 + b^2 + c^2\)? The answer is no because we can’t write

$$ \begin{align*} 7 \neq a^2 + b^2 + c^2. \end{align*} $$

Since if \(a \equiv 7 \bmod 8\), then \(a^2 + b^2 + c^2\) is the sum of integers from the set \(\{0,1,4\}\) only. Note here that we chose squares module 8, because when doing so, we had a very restricted list of \(\{0,1,4\}\). Hence, it is useful to apply module 8 on squares.

Example 5

Can we write any integer \(n\) as a sum of three cubes? so $$ \begin{align*} n = a^3 + b^3 + c^3 \end{align*} $$

Cubes don’t work really well with module 8 but they work really well with module 9. Observe the following

$$ \begin{align*} (a + 3)^3 &\equiv a^3 + 3a^2(3) + 3a(3) + 3^3 \bmod 9 \\ &\equiv a^3 \bmod 9 \end{align*} $$

Observe that adding \(3\) to \(a\) didn’t change its value module 9. In other words

$$ \begin{align*} (a + 3)^3 &\equiv a^3 \bmod 9 \end{align*} $$

So adding \(3\) keeps \(a\) in the same residue class module \(9\). So if \(a = 0\), then

$$ \begin{align*} 0^3 &\equiv 3^3 \equiv 6^3 \equiv 0 \bmod 9 \end{align*} $$

And if \(a = 1\), then

$$ \begin{align*} 1^3 &\equiv 4^3 \equiv 7^3 \equiv 1 \bmod 9 \end{align*} $$

And if \(a = 2\), then

$$ \begin{align*} 2^3 &\equiv 5^3 \equiv 8^3 \equiv -1 \bmod 9 \end{align*} $$

So this means any cube can only leave a remainder of \(\{-1,0,1\}\). Therefore, any sum of three cubes will look like the following module \(9\)

$$ \begin{align*} a^3 + b^3 + c^3 \equiv \{-1,0,1\} + \{-1,0,1\} + \{-1,0,1\} \bmod 9 \end{align*} $$

So now we can just consider all possible sums. For example \(\{(-1-1+0),(-1+0+0), (-1+0+1), (1+1+0)\cdots\)}. In fact, it turns out that the possible sums of cubes can only leave a remainder of

$$ \begin{align*} -3,-2,-1,0,1,2,3 \bmod 9 \end{align*} $$

This means that if \(n \equiv 4 \bmod 9\) or \(n \equiv 5 \bmod 9\), then we don’t have a solution. Only the remainders of \(4\) and \(5\) (module \(9\)) are not possible when writing \(n\) as a sum of three cubes. For example, if \(n = 39\), then \(39 \bmod 9 = 3\) so \(39\) can be written as a sum of three cubes. While \(n = 40\) leaves a remainder of \(40 \bmod 9 = 4\). Therefore, \(40\) can’t be written as a sum of three cubes.

Fermat's Theorem

If \(p\) is prime, \(n \in \mathbb{Z}\), then $$ \begin{align*} n^p \equiv n \bmod p \end{align*} $$

An application of this theorem, suppose we want to show that the following is an integer

$$ \begin{align*} \frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n \end{align*} $$

So multiply by 15

$$ \begin{align*} 3n^5 + 5n^3 + 7n \end{align*} $$

And now, we want to show that this is divisible by 3 and 5 so it’s divisible by 15. To check divisibility by 3, by Fermat’s \(n \equiv n^3 \bmod 3\). Therefore,

$$ \begin{align*} 3n^5 + 5n^3 + 7n &\equiv 3n^5 + 5n + 7n \bmod 3 \\ &\equiv 3n^5 + 12n \bmod 3 \end{align*} $$

Clearly this is divisible by 3. For 5, again by Fermat’s \(n \equiv n^5 \mod 5\)

$$ \begin{align*} 3n^5 + 5n^3 + 7n &\equiv 3n + 5n^3 + 7n \bmod 5 \\ &\equiv 5n^3 + 10n \bmod 5 \end{align*} $$

Clearly this is also divisible by 5. Therefore, \(3n^5 + 5n^3 + 7n\) is divisible by 15 and we are done.

Fermat's Proof

Recall from the previous lecture that \(\binom{p}{k}\) is divisible \(p\) if \(1 \leq k \leq p-1\). This resulted having

$$ \begin{align*} (x+y)^p \equiv x^p + y^p \bmod p \end{align*} $$

So now we prove Fermat’s theorem by induction on \(n\). Note that

$$ \begin{align*} \text{For } n = 0, \quad &0^p \equiv 0 (\bmod p) \\ \text{For } n = 1, \quad &1^p \equiv 1 (\bmod p) \\ \text{For } n = 2, \quad &2^p = (1+1)^p \equiv 1^p + 1^p \equiv 2 (\bmod p) \\ \text{For } n = 3, \quad &3^p = (1+2)^p \equiv 2^p + 1^p \equiv 3 (\bmod p) \end{align*} $$

So now, for the inductive case, suppose that \(n^p \equiv n \bmod p\).

$$ \begin{align*} (n + 1)^p &\equiv n^p + 1^p \quad \\ &\equiv n + 1 \quad \text{(By the inductive hypothesis)} \end{align*} $$

We can do the same thing for negative integers.

Application of Fermat's Theorem

Is \(n=35\) a prime?

It’s obviously not, but hypothetically suppose \(n=35\) has 1000 digits. Then, we can use Fermat’s theorem to say that If 35 is a prime, then the following

$$ \begin{align*} 2^{35} &\equiv 2 \bmod 35 \end{align*} $$

must be true. If this isn’t true, then \(35\) is definitely not a prime. (What if this was true, does that immediately imply that 35 is prime? no, we’ll discuss later). So now how do we calculate \(2^{35}\)? we can do this by reducing the product modulo \(35\) every iteration in order for the product not to get too big. We can start with \(2^6 = 64\) and then keep reducing as follows

$$ \begin{align*} 2^{6} = 64 &\equiv 29 \bmod 35 \\ 2^{7} = 2 \cdot 29 = 58 &\equiv 23 \bmod 35 \\ 2^{8} = 2 \cdot 23 = 46 &\equiv 11 \bmod 35 \end{align*} $$

We can keep doing this until we reach \(2^{35}\). This is good but isn’t great since we have to do this for \(35\) steps until we reach \(2^{35}\). Instead write \(35\) as powers of 2 so

$$ \begin{align*} 35 = 2^5 + 2^1 + 2^0 \end{align*} $$

and now we will square \(2\) in every step as follows

$$ \begin{align*} 2^1 &\equiv 2 \bmod 35 \\ 2^2 &\equiv 4 \bmod 35 \\ 2^4 &\equiv 16 \bmod 35 \\ 2^8 = 16^2 = 256 &\equiv 11 \bmod 35 \\ 2^{16} = 11^2 &\equiv 16 \bmod 35 \\ 2^{32} = 16^2 &\equiv 11 \bmod 35 \\ \end{align*} $$

Thus

$$ \begin{align*} 2^{35} &\equiv 2^{32} \cdot 2^2 \cdot 2^1 \bmod 35 \\ &\equiv 11 \cdot 4 \cdot 2 \bmod 35 \\ &\equiv 88 \bmod 35 \\ &\equiv 18 \bmod 35 \end{align*} $$

So it’s not \(2\). Therefore, \(35\) is not prime. So this is a much faster method. (I believe this is called Modular Exponentiation)

Now, if \(n\) passes the test, It isn’t a guarantee that \(n\) is prime. What we might do is test if \(3^n \equiv 3 \bmod n\). If it still passes, you’ll suspect it is a prime but it’s not a guarantee. It is kind of a probabilistic test at this point. So with some probability, this \(n\) is a prime.

Infinitely Many Primes of the Form \(4n+1\)

Recall from the primes lecture that to prove that we have infinitely many primes of the form \(4n+1\), we used the fact that if a prime divides \(n^2 + 1\), then \(p = 2\) or \(p = 4n+1\) but we never proved this fact. To show, first notice these values for the first \(n\) integers again

\(n\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\)
\(n^2+1\) \(2\) \(5\) \(10\) \(17\) \(26\) \(37\) \(50\) \(65\)
Prime Factors \(2\) \(5\) \(2,5\) \(17\) \(2,13\) \(37\) \(2,5\) \(5,13\)

Observe that all of these prime factors are either 2 or 1 module 4.
Suppose now that \(p\) is an odd prime and it divides \(n^2 + 1\). This means that \(n^2 \equiv -1 \bmod p\). Then we can use Fermat’s theorem to show that \(p\) satisfies \(p \equiv 1 \bmod 4\). So

$$ \begin{align*} n^p &\equiv n \bmod p \\ n^{p-1} &\equiv 1 \bmod p \quad \text{(because $n \not\equiv 0$ and $p$ prime)}\\ (-1)^{(p-1)/2} &\equiv 1 \bmod p \quad \text{(because $n^2 \equiv -1 \bmod p$)}\\ \end{align*} $$

So \((-1)^{(p-1)/2}\) is \(-1\) if \((p-1)/2\) is odd and 1 if \((p-1)/2\) is even. Therefore, we must have \((p-1)/2\) be even. This means that

$$ \begin{align*} (p-1)/2 &\equiv 0 \bmod 2 \\ p-1 &\equiv 0 \bmod 4 \\ p &\equiv 1 \bmod 4 \\ \end{align*} $$

References