Tag Archives: Math 42

Math 42 Spring 2016 Student Showcase

This spring, I taught Math 42: An Introduction to Elementary Number Theory at Brown University. An important aspect of the course was the final project. In these projects, students either followed up on topics that interested them from the semester, or chose and investigated topics related to number theory.  Projects could be done individual or in small groups.

I thought it would be nice to showcase some excellent student projects from my class. Most of the projects were quite good, and some showed extraordinary effort. Some students really dove in and used this as an opportunity to explore and digest a topic far more thoroughly than could possibly be expected from an introductory class such as this one. With the students’ permission, I’ve chosen five student projects (in no particular order) for a blog showcase (impressed by similar sorts  of showcases from Scott Aaronson).

  • Factorization Techniques, by Elvis Nunez and Chris Shaw. In this project, Elvis and Chris look at Fermat Factorization, which looks to factor $n$ by expressing $n = a^2 – b^2$. Further, they investigate improvements to Fermat’s Algorithm by Dixon and Kraitchik. Following this line of investigation leads to the development of the modern quadratic sieve and factor base methods of factorization.

  • Pseudoprimes and Carmichael Numbers, by Emily Riemer. Fermat’s Little Theorem is one of the first “big idea” theorems we encounter in the course, and we came back to it again and again throughout. Emily explored the Fermat’s Little Theorem as a primality test, leading to pseudoprimes, strong pseudoprimes, and Carmichael numbers. [As an aside, one of her references concerning Carmichael numbers were notes from an algebraic number theory class taught by Matt Baker, who first got me interested in number theory].

  • Continued Fractions and Pell’s Equation, by Max Lahn and Jonathan Spiegel. As it happened, I did not have time to teach continued fractions in the course.  So Max and Jonathan decided to look at them on their own. They explore some ideas related to the convergence of continued fractions and see how one uses continued fractions to solve Pell’s Equation.

  • Quantum Computing, by Edward Hu and Chris Long. Edward and Chris explore quantum computing with particular emphasis towards gaining some idea of how Shor’s factorization algorithm works. For some of the more complicated ideas, like the quantum Fourier transform, they make use of heuristic and analogy to purvey the main ideas.

  • Fermat’s Last Theorem, by Dylan Groos, Natalie Schudrowitz, and Kenneth Berglund. Dylan, Natalie, and Kenneth provide a historical look at attacks on Fermat’s Last Theorem. They examine proofs for $n=4$ and Sophie Germaine’s remarkable advances. They also touch on elliptic curves and modular forms, hinting at some of the deep ideas lying beneath the surface.

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Math 420: Supplement on Gaussian Integers II

This is a secondary supplemental note on the Gaussian integers, written for my Spring 2016 Elementary Number Theory Class at Brown University. This note is also available as a pdf document.

In this note, we cover the following topics.

  1. Assumed prerequisites from other lectures.
  2. Which regular integer primes are sums of squares?
  3. How can we classify all Gaussian primes?

1. Assumed Prerequisites

Although this note comes shortly after the previous note on the Gaussian integers, we covered some material from the book in the middle. In particular, we will assume use the results from chapters 20 and 21 from the textbook.

Most importantly, for $latex {p}$ a prime and $latex {a}$ an integer not divisible by $latex {p}$, recall the Legendre symbol $latex {\left(\frac{a}{p}\right)}$, which is defined to be $latex {1}$ if $latex {a}$ is a square mod $latex {p}$ and $latex {-1}$ if $latex {a}$ is not a square mod $latex {p}$. Then we have shown Euler’s Criterion, which states that

$$ a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right) \pmod p, \tag{1}$$
and which gives a very efficient way of determining whether a given number $latex {a}$ is a square mod $latex {p}$.

We used Euler’s Criterion to find out exactly when $latex {-1}$ is a square mod $latex {p}$. In particular, we concluded that for each odd prime $latex {p}$, we have

$$ \left(\frac{-1}{p}\right) = \begin{cases} 1 & \text{ if } p \equiv 1 \pmod 4 \ -1 & \text{ if } p \equiv 3 \pmod 4 \end{cases}. \tag{2}$$
Finally, we assume familiarity with the notation and ideas from the previous note on the Gaussian integers.

2. Understanding When $latex {p = a^2 + b^2}$.

Throughout this section, $latex {p}$ will be a normal odd prime. The case $latex {p = 2}$ is a bit different, and we will need to handle it separately. When used, the letters $latex {a}$ and $latex {b}$ will denote normal integers, and $latex {q_1,q_2}$ will denote Gaussian integers.

We will be looking at the following four statements.

  1. $latex {p \equiv 1 \pmod 4}$
  2. $latex {\left(\frac{-1}{p}\right) = 1}$
  3. $latex {p}$ is not a Gaussian prime
  4. $latex {p = a^2 + b^2}$

Our goal will be to show that each of these statements are equivalent. In order to show this, we will show that

$$ (1) \implies (2) \implies (3) \implies (4) \implies (1). \tag{3}$$
Do you see why this means that they are all equivalent?

This naturally breaks down into four lemmas.

We have actually already shown one.

Lemma 1 $latex {(1) \implies (2)}$.

Proof: We have already proved this claim! This is exactly what we get from Euler’s Criterion applied to $latex {-1}$, as mentioned in the first section. $latex \Box$

There is one more that is somewhat straightfoward, and which does not rely on going up to the Gaussian integers.

Lemma 2 $latex {(4) \implies (1)}$.

Proof: We have an odd prime $latex {p}$ which is a sum of squares $latex {p = a^2 + b^2}$. If we look mod $latex {4}$, we are led to consider $$ p = a^2 + b^2 \pmod 4. \tag{4}$$
What are the possible values of $latex {a^2 \pmod 4}$? A quick check shows that the only possibilites are $latex {a^2 \equiv 0, 1 \pmod 4}$.

So what are the possible values of $latex {a^2 + b^2 \pmod 4}$? We must have one of $latex {p \equiv 0, 1, 2 \pmod 4}$. Clearly, we cannot have $latex {p \equiv 0 \pmod 4}$, as then $latex {4 \mid p}$. Similarly, we cannot have $latex {p \equiv 2 \pmod 4}$, as then $latex {2 \mid p}$. So we necessarily have $latex {p \equiv 1 \pmod 4}$, which is what we were trying to prove. $latex \Box$

For the remaining two pieces, we will dive into the Gaussian integers.

Lemma 3 $latex {(2) \implies (3)}$.

Proof: As $latex {\left(\frac{-1}{p}\right) = 1}$, we know there is some $latex {a}$ so that $latex {a^2 \equiv -1 \pmod p}$. Rearranging, this becomes $latex {a^2 + 1 \equiv 0 \pmod p}$.

Over the normal integers, we are at an impasse, as all this tells us is that $latex {p \mid (a^2 + 1)}$. But if we suddenly view this within the Gaussian integers, then $latex {a^2 + 1}$ factors as $latex {a^2 + 1 = (a + i)(a – i)}$.

So we have that $latex {p \mid (a+i)(a-i)}$. If $latex {p}$ were a Gaussian prime, then we would necessarily have $latex {p \mid (a+i)}$ or $latex {p \mid (a-i)}$. (Do you see why?)

But is it true that $latex {p}$ divides $latex {a + i}$ or $latex {a – i}$? For instance, does $latex {p}$ divide $latex {a + i}$? No! If so, then $latex {\frac{a}{p} + \frac{i}{p}}$ would be a Gaussian integer, which is clearly not true.

So $latex {p}$ does not divide $latex {a + i}$ or $latex {a-i}$, and we must therefore conclude that $latex {p}$ is not a Gaussian prime. $latex \Box$

Lemma 4 $latex {(3) \implies (4)}$.

Proof: We now know that $latex {p}$ is not a Gaussian prime. In particular, this means that $latex {p}$ is not irreducible, and so it has a nontrivial factorization in the Gaussian integers. (For example, $latex {5}$ is a regular prime, but it is not a Gaussian prime. It factors as $latex {5 = (1 + 2i)(1 – 2i)}$ in the Gaussian integers.)

Let’s denote this nontrivial factorization as $latex {p = q_1 q_2}$. By nontrivial, we mean that neither $latex {q_1}$ nor $latex {q_2}$ are units, i.e. $latex {N(q_1), N(q_2) > 1}$. Taking norms, we see that $latex {N(p) = N(q_1) N(q_2)}$.

We can evaluate $latex {N(p) = p^2}$, so we have that $latex {p^2 = N(q_1) N(q_2)}$. Both $latex {N(q_1)}$ and $latex {N(q_2)}$ are integers, and their product is $latex {p^2}$. Yet $latex {p^2}$ has exactly two different factorizations: $latex {p^2 = 1 \cdot p^2 = p \cdot p}$. Since $latex {N(q_1), N(q_2) > 1}$, we must have the latter.

So we see that $latex {N(q_1) = N(q_2) = p}$. As $latex {q_1, q_2}$ are Gaussian integers, we can write $latex {q_1 = a + bi}$ for some $latex {a, b}$. Then since $latex {N(q_1) = p}$, we see that $latex {N(q_1) = a^2 + b^2}$. And so $latex {p}$ is a sum of squares, ending the proof. $latex \Box$

Notice that $latex {2 = 1 + 1}$ is also a sum of squares. Then all together, we can say the following theorem.

Theorem 5 A regular prime $latex {p}$ can be written as a sum of two squares, $$ p = a^2 + b^2, \tag{5}$$
exactly when $latex {p = 2}$ or $latex {p \equiv 1 \pmod 4}$.

A remarkable aspect of this theorem is that it is entirely a statement about the behaviour of the regular integers. Yet in our proof, we used the Gaussian integers in a very fundamental way. Isn’t that strange?

You might notice that in the textbook, Dr. Silverman presents a proof that does not rely on the Gaussian integers. While interesting and clever, I find that the proof using the Gaussian integers better illustrates the deep connections between and around the structures we have been studying in this course so far. Everything connects!

Example 1 The prime $latex {5}$ is $latex {1 \pmod 4}$, and so $latex {5}$ is a sum of squares. In particular, $latex {5 = 1^2 + 2^2}$.

Example 2 The prime $latex {101}$ is $latex {1 \pmod 4}$, and so is a sum of squares. Our proof is not constructive, so a priori we do not know what squares sum to $latex {101}$. But in this case, we see that $latex {101 = 1^2 + 10^2}$.

Example 3 The prime $latex {97}$ is $latex {1 \pmod 4}$, and so it also a sum of squares. It’s less obvious what the squares are in this case. It turns out that $latex {97 = 4^2 + 9^2}$.

Example 4 The prime $latex {43}$ is $latex {3 \pmod 4}$, and so is not a sum of squares.

3. Classification of Gaussian Primes

In the previous section, we showed that each integer prime $latex {p \equiv 1 \pmod 4}$ actually splits into a product of two Gaussian numbers $latex {q_1}$ and $latex {q_2}$. In fact, since $latex {N(q_1) = p}$ is a regular prime, $latex {q_1}$ is a Gaussian irreducible and therefore a Gaussian prime (can you prove this? This is a nice midterm question.)

So in fact, $latex {p \equiv 1 \pmod 4}$ splits in to the product of two Gaussian primes $latex {q_1}$ and $latex {q_2}$.

In this way, we’ve found infinitely many Gaussian primes. Take a regular prime congruent to $latex {1 \pmod 4}$. Then we know that it splits into two Gaussian primes. Further, if we know how to write $latex {p = a^2 + b^2}$, then we know that $latex {q_1 = a + bi}$ and $latex {q_2 = a – bi}$ are those two Gaussian primes.

In general, we will find all Gaussian primes by determining their interaction with regular primes.

Suppose $latex {q}$ is a Gaussian prime. Then on the one hand, $latex {N(q) = q \overline{q}}$. On the other hand, $latex {N(q) = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}}$ is some regular integer. Since $latex {q}$ is a Gaussian prime (and so $latex {q \mid w_1 w_2}$ means that $latex {q \mid w_1}$ or $latex {q \mid w_2}$), we know that $latex {q \mid p_j}$ for some regular integer prime $latex {p_j}$.

So one way to classify Gaussian primes is to look at every regular integer prime and see which Gaussian primes divide it. We have figured this out for all primes $latex {p \equiv 1 \pmod 4}$. We can handle $latex {2}$ by noticing that $latex {2 = (1 + i) (1-i)}$. Both $latex {(1+i)}$ and $latex {(1-i)}$ are Gaussian primes.

The only primes left are those regular primes with $latex {p \equiv 3 \pmod 4}$. We actually already covered the key idea in the previous section.

Lemma 6 If $latex {p \equiv 3 \pmod 4}$ is a regular prime, then $latex {p}$ is also a Gaussian prime.

Proof: In the previous section, we showed that if $latex {p}$ is not a Gaussian prime, then $latex {p = a^2 + b^2}$ for some integers $latex {a,b}$, and then $latex { p \equiv 1 \pmod 4}$. Since $latex {p \not \equiv 1 \pmod 4}$, we see that $latex {p}$ is a Gaussian prime. $latex \Box$

In total, we have classified all Gaussian primes.

Theorem 7 The Gaussian primes are given by

  1. $latex {(1+i), (1-i)}$
  2. Regular primes $latex {p \equiv 3 \pmod 4}$
  3. The factors $latex {q_1 q_2}$ of a regular prime $latex {p \equiv 1 \pmod 4}$. Further, these primes are given by $latex {a \pm bi}$, where $latex {p = a^2 + b^2}$.

 

4. Concluding Remarks

I hope that it’s clear that the regular integers and the Gaussian integers are deeply connected and intertwined. Number theoretic questions in one constantly lead us to investigate the other. As one dives deeper into number theory, more and different integer-like rings appear, all deeply connected.

Each time I teach the Gaussian integers, I cannot help but feel the sense that this is a hint at a deep structural understanding of what is really going on. The interplay between the Gaussian integers and the regular integers is one of my favorite aspects of elementary number theory, which is one reason why I deviated so strongly from the textbook to include it. I hope you enjoyed it too.

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Math 420: Supplement on Gaussian Integers

This is a brief supplemental note on the Gaussian integers, written for my Spring 2016 Elementary Number Class at Brown University. With respect to the book, the nearest material is the material in Chapters 35 and 36, but we take a very different approach.

A pdf of this note can be found here. I’m sure there are typos, so feel free to ask me or correct me if you think something is amiss.

In this note, we cover the following topics.

  1. What are the Gaussian integers?
  2. Unique factorization within the Gaussian integers.
  3. An application of the Gaussian integers to the Diophantine equation $latex {y^2 = x^3 – 1}$.
  4. Other integer-like sets: general rings.
  5. Specific examples within $latex {\mathbb{Z}[\sqrt{2}]}$ and $latex {\mathbb{Z}[\sqrt{-5}]}$.

1. What are the Gaussian Integers?

The Gaussian Integers are the set of numbers of the form $latex {a + bi}$, where $latex {a}$ and $latex {b}$ are normal integers and $latex {i}$ is a number satisfying $latex {i^2 = -1}$. As a collection, the Gaussian Integers are represented by the symbol $latex {\mathbb{Z}[i]}$, or sometimes $latex {\mathbb{Z}[\sqrt{-1}]}$. These might be pronounced either as The Gaussian Integers or as Z append i.

In many ways, the Gaussian integers behave very much like the regular integers. We’ve been studying the qualities of the integers, but we should ask — which properties are really properties of the integers, and which properties hold in greater generality? Is it the integers themselves that are special, or is there something bigger and deeper going on?

These are the main questions that we ask and make some progress towards in these notes. But first, we need to describe some properties of Gaussian integers.

We will usually use the symbols $latex {z = a + bi}$ to represent our typical Gaussian integer. One adds and multiples two Gaussian integers just as you would add and multiply two complex numbers. Informally, you treat $latex {i}$ like a polynomial indeterminate $latex {X}$, except that it satisfies the relation $latex {X^2 = -1}$.

Definition 1 For each complex number $latex {z = a + bi}$, we define the conjugate of $latex {z}$, written as $latex {\overline{z}}$, by
\begin{equation}
\overline{z} = a – bi.
\end{equation}
We also define the norm of $latex {z}$, written as $latex {N(z)}$, by
\begin{equation}
N(z) = a^2 + b^2.
\end{equation}

You can check that $latex {N(z) = z \overline{z}}$ (and in fact this is one of your assigned problems). You can also chack that $latex {N(zw) = N(z)N(w)}$, or rather that the norm is multiplicative (this is also one of your assigned problems).

Even from our notation, it’s intuitive that $latex {z = a + bi}$ has two parts, the part corresponding to $latex {a}$ and the part corresponding to $latex {b}$. We call $latex {a}$ the real part of $latex {z}$, written as $latex {\Re z = a}$, and we call $latex {b}$ the imaginary part of $latex {z}$, written as $latex {\Im z = b}$. I should add that the name ”imaginary number” is a poor name that reflects historical reluctance to view complex numbers as acceptable. For that matter, the name ”complex number” is also a poor name.

As a brief example, consider the Gaussian integer $latex {z = 2 + 5i}$. Then $latex {N(z) = 4 + 25 = 29}$, $latex {\Re z = 2}$, $latex {\Im z = 5}$, and $latex {\overline{z} = 2 – 5i}$.

We can ask similar questions to those we asked about the regular integers. What does it mean for $latex {z \mid w}$ in the complex case?

Definition 2 We say that a Gaussian integer $latex {z}$ divides another Gaussian integer $latex {w}$ if there is some Gaussian integer $latex {k}$ so that $latex {zk = w}$. In this case, we write $latex {z \mid w}$, just as we write for regular integers.

For the integers, we immediately began to study the properties of the primes, which in many ways were the building blocks of the integers. Recall that for the regular integers, we said $latex {p}$ was a prime if its only divisors were $latex {\pm 1}$ and $latex {\pm p}$. In the Gaussian integers, the four numbers $latex {\pm 1, \pm i}$ play the same role as $latex {\pm 1}$ in the usual integers. These four numbers are distinguished as being the only four Gaussian integers with norm equal to $latex {1}$.

That is, the only solutions to $latex {N(z) = 1}$ where $latex {z}$ is a Gaussian integer are $latex {z = \pm 1, \pm i}$. We call these four numbers the Gaussian units.

With this in mind, we are ready to define the notion of a prime for the Gaussian integers.

Definition 3 We say that a Gaussian integer $latex {z}$ with $latex {N(z) > 1}$ is a Gaussian prime if the only divisors of $latex {z}$ are $latex {u}$ and $latex {uz}$, where $latex {u = \pm 1, \pm i}$ is a Gaussian unit.

Remark 1 When we look at other integer-like sets, we will actually use a different definition of a prime.

It’s natural to ask whether the normal primes in $latex {\mathbb{Z}}$ are also primes in $latex {\mathbb{Z}[i]}$. And the answer is no. For instance, $latex {5}$ is a prime in $latex {\mathbb{Z}}$, but
\begin{equation}
5 = (1 + 2i)(1 – 2i)
\end{equation}
in the Gaussian integers. However, the two Gaussian integers $latex {1 + 4i}$ and $latex {1 – 4i}$ are prime. It also happens to be that $latex {3}$ is a Gaussian prime. We will continue to investigate which numbers are Gaussian primes over the next few lectures.

With a concept of a prime, it’s also natural to ask whether or not the primes form the building blocks for the Gaussian integers like they form the building blocks for the regular integers. We take up this in our next topic.

2. Unique Factorization in the Gaussian Integers

Let us review the steps that we followed to prove unique factorization for $latex {\mathbb{Z}}$.

  1. We proved that for $latex {a,b}$ in $latex {\mathbb{Z}}$ with $latex {b \neq 0}$, there exist unique $latex {q}$ and $latex {r}$ such that $latex {a = bq + r}$ with $latex {0 \leq r < b}$. This is called the Division Algorithm.
  2. By repeatedly applying the Division Algorithm, we proved the Euclidean Algorithm. In particular, we showed that the last nonzero remainder was the GCD of our initial numbers.
  3. By performing reverse substition on the steps of the Euclidean Algorithm, we showed that there are integer solutions in $latex {x,y}$ to the Diophantine equation $latex {ax + by = \gcd(a,b)}$. This is often called Bezout’s Theorem or Bezout’s Lemma, although we never called it by that name in class.
  4. With Bezout’s Theorem, we showed that if a prime $latex {p}$ divides $latex {ab}$, then $latex {p \mid a}$ or $latex {p \mid b}$. This is the crucial step towards proving Unique Factorization.
  5. We then proved Unique Factorization.

Each step of this process can be repeated for the Gaussian integers, with a few notable differences. Remarkably, once we have the division algorithm, each proof is almost identical for $latex {\mathbb{Z}[i]}$ as it is for $latex {\mathbb{Z}}$. So we will prove the division algorithm, and then give sketches of the remaining ideas, highlighting the differences that come up along the way.

In the division algorithm, we require the remainder $latex {r}$ to ”be less than what we are dividing by.” A big problem in translating this to the Gaussian integers is that the Gaussian integers are not ordered. That is, we don’t have a concept of being greater than or less than for $latex {\mathbb{Z}[i]}$.

When this sort of problem emerges, we will get around this by taking norms. Since the norm of a Gaussian integer is a typical integer, we will be able to use the ordering of the integers to order our norms.

Theorem 4 For $latex {z,w}$ in $latex {\mathbb{Z}[i]}$ with $latex {w \neq 0}$, there exist $latex {q}$ and $latex {r}$ in $latex {\mathbb{Z}[i]}$ such that $latex {z = qw + r}$ with $latex {N(r) < N(w)}$.

Proof: Here, we will cheat a little bit and use properties about general complex numbers and the rationals to perform this proof. One can give an entirely intrinsic proof, but I like the approach I give as it also informs how to actually compute the $latex {q}$ and $latex {r}$.

The entire proof boils down to the idea of writing $latex {z/w}$ as a fraction and approximating the real and imaginary parts by the nearest integers.

Let us now transcribe that idea. We will need to introduce some additional symbols. Let $latex {z = a_1 + b_1 i}$ and $latex {w = a_2 + b_2 i}$.

Then
\begin{align}
\frac{z}{w} &= \frac{a_1 + b_1 i}{a_2 + b_2 i} = \frac{a_1 + b_1 i}{a_2 + b_2 i} \frac{a_2 – b_2 i}{a_2 – b_2 i} \\
&= \frac{a_1a_2 + b_1 b_2}{a_2^2 + b_2^2} + i \frac{b_1 a_2 – a_1 b_2}{a_2^2 + b_2 ^2} \\
&= u + iv.
\end{align}
By rationalizing the denominator by multiplying by $latex {\overline{w}/ \overline{w}}$, we are able to separate out the real and imaginary parts. In this final expression, we have named $latex {u}$ to be the real part and $latex {v}$ to be the imaginary part. Notice that $latex {u}$ and $latex {v}$ are normal rational numbers.

We know that for any rational number $latex {u}$, there is an integer $latex {u’}$ such that $latex {\lvert u – u’ \rvert \leq \frac{1}{2}}$. Let $latex {u’}$ and $latex {v’}$ be integers within $latex {1/2}$ of $latex {u}$ and $latex {v}$ above, respectively.

Then we claim that we can choose $latex {q = u’ + i v’}$ to be the $latex {q}$ in the theorem statement, and let $latex {r}$ be the resulting remainder, $latex {r = z – qw}$. We need to check that $latex {N(r) < N(w)}$. We will check that explicitly.

We compute
\begin{align}
N(r) &= N(z – qw) = N\left(w \left(\frac{z}{w} – q\right)\right) = N(w) N\left(\frac{z}{w} – q\right).
\end{align}
Note that we have used that $latex {N(ab) = N(a)N(b)}$. In this final expression, we have already come across $latex {\frac{z}{w}}$ before — it’s exactly what we called $latex {u + iv}$. And we called $latex {q = u’ + i v’}$. So our final expression is the same as
\begin{equation}
N(r) = N(w) N(u + iv – u’ – i v’) = N(w) N\left( (u – u’) + i (v – v’)\right).
\end{equation}
How large can the real and imaginary parts of $latex {(u-u’) + i (v – v’)}$ be? By our choice of $latex {u’}$ and $latex {v’}$, they can be at most $latex {1/2}$.

So we have that
\begin{equation}
N(r) \leq N(w) N\left( (\tfrac{1}{2})^2 + (\tfrac{1}{2})^2\right) = \frac{1}{2} N(w).
\end{equation}
And so in particular, we have that $latex {N(r) < N(w)}$ as we needed. $latex \Box$

Note that in this proof, we did not actually show that $latex {q}$ or $latex {r}$ are unique. In fact, unlike the case in the regular integers, it is not true that $latex {q}$ and $latex {r}$ are unique.

Example 1 Consider $latex {3+5i, 1 + 2i}$. Then we compute
\begin{equation}
\frac{3+5i}{1+2i} = \frac{3+5i}{1+2i}\frac{1-2i}{1-2i} = \frac{13}{5} + i \frac{-1}{5}.
\end{equation}
The closest integer to $latex {13/5}$ is $latex {3}$, and the closest integer to $latex {-1/5}$ is $latex {0}$. So we take $latex {q = 3}$. Then $latex {r = (3+5i) – (1+2i)3 = -i}$, and we see in total that
\begin{equation}
3+5i = (1+2i) 3 – i.
\end{equation}
Note that $latex {N(-i) = 1}$ and $latex {N(1 + 2i) = 5}$, so this choice of $latex {q}$ and $latex {r}$ works.

As $latex {13/5}$ is sort of close to $latex {2}$, what if we chose $latex {q = 2}$ instead? Then $latex {r = (3 + 5i) – (1 + 2i)2 = 1 + i}$, leading to the overall expression
\begin{equation}
3_5i = (1 + 2i) 2 + (1 + i).
\end{equation}
Note that $latex {N(1+i) = 2 < N(1+2i) = 5}$, so that this choice of $latex {q}$ and $latex {r}$ also works.

This is an example of how the choice of $latex {q}$ and $latex {r}$ is not well-defined for the Gaussian integers. In fact, even if one decides to choose $latex {q}$ to that $latex {N(r)}$ is minimal, the resulting choices are still not necessarily unique.

This may come as a surprise. The letters $latex {q}$ and $latex {r}$ come from our tendency to call those numbers the quotient and remainder after division. We have shown that the quotient and remainder are not well-defined, so it does not make sense to talk about ”the remainder” or ”the quotient.” This is a bit strange!

Are we able to prove unique factorization when the process of division itself seems to lead to ambiguities? Let us proceed forwards and try to see.

Our next goal is to prove the Euclidean Algorithm. By this, we mean that by repeatedly performing the division algorithm starting with two Gaussian integers $latex {z}$ and $latex {w}$, we hope to get a sequence of remainders with the last nonzero remainder giving a greatest common divisor of $latex {z}$ and $latex {w}$.

Before we can do that, we need to ask a much more basic question. What do we mean by a greatest common divisor? In particular, the Gaussian integers are not ordered, so it does not make sense to say whether one Gaussian integer is bigger than another.

For instance, is it true that $latex {i > 1}$? If so, then certainly $latex {i}$ is positive. We know that multiplying both sides of an inequality by a positive number doesn’t change that inequality. So multiplying $latex {i > 1}$ by $latex {i}$ leads to $latex {-1 > i}$, which is absurd if $latex {i}$ was supposed to be positive!

To remedy this problem, we will choose a common divisor of $latex {z}$ and $latex {w}$ with the greatest norm (which makes sense, as the norm is a regular integer and thus is well-ordered). But the problem here, just as with the division algorithm, is that there may or may not be multiple such numbers. So we cannot talk about ”the greatest common divisor” and instead talk about ”a greatest common divisor.” To paraphrase Lewis Carroll’s\footnote{Carroll was also a mathematician, and hid some nice mathematics inside some of his works.} Alice, things are getting curiouser and curiouser!

Definition 5 For nonzero $latex {z,w}$ in $latex {\mathbb{Z}[i]}$, a greatest common divisor of $latex {z}$ and $latex {w}$, denoted by $latex {\gcd(z,w)}$, is a common divisor with largest norm. That is, if $latex {c}$ is another common divisor of $latex {z}$ and $latex {w}$, then $latex {N(c) \leq N(\gcd(z,w))}$.

If $latex {N(\gcd(z,w)) = 1}$, then we say that $latex {z}$ and $latex {w}$ are relatively prime. Said differently, if $latex {1}$ is a greatest common divisor of $latex {z}$ and $latex {w}$, then we say that $latex {z}$ and $latex {w}$ are relatively prime.

Remark 2 Note that $latex {\gcd(z,w)}$ as we’re writing it is not actually well-defined, and may stand for any greatest common divisor of $latex {z}$ and $latex {w}$.

With this definition in mind, the proof of the Euclidean Algorithm is almost identical to the proof of the Euclidean Algorithm for the regular integers. As with the regular integers, we need the following result, which we will use over and over again.

Lemma 6 Suppose that $latex {z \mid w_1}$ and $latex {z \mid w_2}$. Then for any $latex {x,y}$ in $latex {\mathbb{Z}[i]}$, we have that $latex {z \mid (x w_1 + y w_2)}$.

Proof: As $latex {z \mid w_1}$, there is some Gaussian integer $latex {k_1}$ such that $latex {z k_1 = w_1}$. Similarly, there is some Gaussian integer $latex {k_2}$ such that $latex {z k_2 = w_2}$.

Then $latex {xw_1 + yw_2 = zxk_1 + zyk_2 = z(xk_1 + yk_2)}$, which is divisible by $latex {z}$ as this is the definition of divisibility. $latex \Box$

Notice that this proof is identical to the analogous statement in the integers, except with differently chosen symbols. That is how the proof of the Euclidean Algorithm goes as well.

Theorem 7 let $latex {z,w}$ be nonzero Gaussian integers. Recursively apply the division algorithm, starting with the pair $latex {z, w}$ and then choosing the quotient and remainder in one equation the new pair for the next. The last nonzero remainder is divisible by all common divisors of $latex {z,w}$, is itself a common divisor, and so the last nonzero remainder is a greatest common divisor of $latex {z}$ and $latex {w}$.

Symbolically, this looks like
\begin{align}
z &= q_1 w + r_1, \quad N(r_1) < N(w) \\\\
w &= q_2 r_1 + r_2, \quad N(r_2) < N(r_1) \\\\
r_1 &= q_3 r_2 + r_3, \quad N(r_3) < N(r_2) \\\\
\cdots &= \cdots \\\\
r_k &= q_{k+2} r_{k+1} + r_{k+2}, \quad N(r_{k+2}) < N(r_{k+1}) \\\\
r_{k+1} &= q_{k+3} r_{k+2} + 0,
\end{align}
where $latex {r_{k+2}}$ is the last nonzero remainder, which we claim is a greatest common divisor of $latex {z}$ and $latex {w}$.

Proof: We are claiming several thing. Firstly, we should prove our implicit claim that this algorithm terminates at all. Is it obvious that we should eventually reach a zero remainder?

In order to see this, we look at the norms of the remainders. After each step in the algorithm, the norm of the remainder is smaller than the previous step. As the norms are always nonnegative integers, and we know there does not exist an infinite list of decreasing positive integers, we see that the list of nonzero remainders is finite. So the algorithm terminates.

We now want to prove that the last nonzero remainder is a common divisor and is in fact a greatest common divisor. The proof is actually identical to the proof in the integer case, merely with a different choice of symbols.

Here, we only sketch the argument. Then the rest of the argument can be found by comparing with the proof of the Euclidean Algorithm for $latex {\mathbb{Z}}$ as found in the course textbook.

For ease of exposition, suppose that the algorithm terminated in exatly 3 steps, so that we have
\begin{align}
z &= q_1 w + r_1, \\
w &= q_2 r_1 + r_2 \\
r_1 &= q_3 r_2 + 0.
\end{align}

On the one hand, suppose that $latex {d}$ is a common divisor of $latex {z}$ and $latex {w}$. Then by our previous lemma, $latex {d \mid z – q_1 w = r_1}$, so that we see that $latex {d}$ is a divisor of $latex {r_1}$ as well. Applying to the next line, we have that $latex {d \mid w}$ and $latex {d \mid r_1}$, so that $latex {d \mid w – q_2 r_1 = r_2}$. So every common divisor of $latex {z}$ and $latex {w}$ is a divisor of the last nonzero remainder $latex {r_2}$.

On the other hand, $latex {r_2 \mid r_1}$ by the last line of the algorithm. Then as $latex {r_2 \mid r_1}$ and $latex {r_2 \mid r_1}$, we know that $latex {r_2 \mid q_2 r_1 + r_2 = w}$. Applying this to the first line, as $latex {r_2 \mid r_1}$ and $latex {r_2 \mid w}$, we know that $latex {r_2 \mid q_1 w + r_1 = z}$. So $latex {r_2}$ is a common divisor.

We have shown that $latex {r_2}$ is a common divisor of $latex {z}$ and $latex {w}$, and that every common divisor of $latex {z}$ and $latex {w}$ divides $latex {r_2}$. How do we show that $latex {r_2}$ is a greatest common divisor?

Suppose that $latex {d}$ is a common divisor of $latex {z}$ and $latex {w}$, so that we know that $latex {d \mid r_2}$. In particular, this means that there is some nonzero $latex {k}$ so that $latex {dk = r_2}$. Taking norms, this means that $latex {N(dk) = N(d)N(k) = N(r_2)}$. As $latex {N(d)}$ and $latex {N(k)}$ are both at least $latex {1}$, this means that $latex {N(d) \leq N(r_2)}$.

This is true for every common divisor $latex {d}$, and so $latex {N(r_2)}$ is at least as large as the norm of any common divisor of $latex {z}$ and $latex {w}$. Thus $latex {r_2}$ is a greatest common divisor.

The argument carries on in the same way for when there are more steps in the algorithm. $latex \Box$

Theorem 8 The greatest common divisor of $latex {z}$ and $latex {w}$ is well-defined, up to multiplication by $latex {\pm 1, \pm i}$. In other words, if $latex {\gcd(z,w)}$ is a greatest common divisor of $latex {z}$ and $latex {w}$, then all greatest common divisors of $latex {z}$ and $latex {w}$ are given by $latex {\pm \gcd(z,w), \pm i \gcd(z,w)}$.

Proof: Suppose $latex {d}$ is a greatest common divisor, and let $latex {\gcd(z,w)}$ denote a greatest common divisor resulting from an application of the Euclidean Algorithm. Then we know that $latex {d \mid \gcd(z,w)}$, so that there is some $latex {k}$ so that $latex {dk = \gcd(z,w)}$. Taking norms, we see that $latex {N(d)N(k) = N(\gcd(z,w)}$.

But as both $latex {d}$ and $latex {\gcd(z,w)}$ are greatest common divisors, we must have that $latex {N(d) = N(\gcd(z,w))}$. So $latex {N(k) = 1}$. The only Gaussian integers with norm one are $latex {\pm 1, \pm i}$, so we have that $latex {du = \gcd(z,w)}$ where $latex {u}$ is one of the four Gaussian units, $latex {\pm 1, \pm i}$.

Conversely, it’s clear that the four numbers $latex {\pm \gcd(z,w), \pm i \gcd(z,w)}$ are all greatest common divisors. $latex \Box$

Now that we have the Euclidean Algorithm, we can go towards unique factorization in $latex {\mathbb{Z}[i]}$. Let $latex {g}$ denote a greatest common divisor of $latex {z}$ and $latex {w}$. Reverse substitution in the Euclidean Algorithm shows that we can find Gaussian integer solutions $latex {x,y}$ to the (complex) linear Diophantine equation
\begin{equation}
zx + wy = g.
\end{equation}
Let’s see an example.

Example 2 Consider $latex {32 + 9i}$ and $latex {4 + 11i}$. The Euclidean Algorithm looks like
\begin{align}
32 + 9i &= (4 + 11i)(2 – 2i) + 2 – 5i, \\\\
4 + 11i &= (2 – 5i)(-2 + i) + 3 – i, \\\\
2 – 5i &= (3-i)(1-i) – i, \\\\
3 – i &= -i (1 + 3i) + 0.
\end{align}
So we know that $latex {-i}$ is a greatest common divisor of $latex {32 + 9i}$ and $latex {4 + 11i}$, and so we know that $latex {32+9i}$ and $latex {4 + 11i}$ are relatively prime. Let us try to find a solution to the Diophantine equation
\begin{equation}
x(32 + 9i) + y(4 + 11i) = 1.
\end{equation}
Performing reverse substition, we see that
\begin{align}
-i &= (2 – 5i) – (3-i)(1-i) \\\\
&= (2 – 5i) – (4 + 11i – (2-5i)(-2 + i))(1-i) \\\\
&= (2 – 5i) – (4 + 11i)(1 – i) + (2 – 5i)(-2 + 1)(1 – i) \\\\
&= (2 – 5i)(3i) – (4 + 11i)(1 – i) \\\\
&= (32 + 9i – (4 + 11i)(2 – 2i))(3i) – (4 + 11i)(1 – i) \\\\
&= (32 + 9i) 3i – (4 + 11i)(2 – 2i)(3i) – (4 + 11i)(1-i) \\\\
&= (32 + 9i) 3i – (4 + 11i)(7 + 5i).
\end{align}
Multiplying this through by $latex {i}$, we have that
\begin{equation}
1 = (32 + 9i) (-3) + (4 + 11i)(5 – 7i).
\end{equation}
So one solution is $latex {(x,y) = (-3, 5 – 7i)}$.

Although this looks more complicated, the process is the same as in the case over the regular integers. The apparent higher difficulty comes mostly from our lack of familiarity with basic arithmetic in $latex {\mathbb{Z}[i]}$.

The rest of the argument is now exactly as in the integers.

Theorem 9 Suppose that $latex {z, w}$ are relatively prime, and that $latex {z \mid wv}$. Then $latex {z \mid v}$.

Proof: This is left as an exercise (and will appear on the next midterm in some form — cheers to you if you’ve read this far in these notes). But it’s now the almost the same as in the regular integers. $latex \Box$

Theorem 10 Let $latex {z}$ be a Gaussian integer with $latex {N(z) > 1}$. Then $latex {z}$ can be written uniquely as a product of Gaussian primes, up to multiplication by one of the Gaussian units $latex {\pm 1, \pm i}$.

Proof: We only sketch part of the proof. There are multiple ways of doing this, but we present the one most similar to what we’ve done for the integers. If there are Gaussian integers without unique factorization, then there are some (maybe they tie) with minimal norm. So let $latex {z}$ be a Gaussian integer of minimal norm without unique factorization. Then we can write
\begin{equation}
p_1 p_2 \cdots p_k = z = q_1 q_2 \cdots q_\ell,
\end{equation}
where the $latex {p}$ and $latex {q}$ are all primes. As $latex {p_1 \mid z = q_1 q_2 \cdots q_\ell}$, we know that $latex {p_1}$ divides one of the $latex {q}$ (by Theorem~9), and so (up to units) we can say that $latex {p_1}$ is one of the $latex {q}$ primes. We can divide each side by $latex {p_1}$ and we get two supposedly different factorizations of a Gaussian integer of norm $latex {N(z)/N(p_1) < N(z)}$, which is less than the least norm of an integer without unique factorization (by what we supposed). This is a contradiction, and we can conclude that there are no Gaussian integers without unique factorization. $latex \Box$

If this seems unclear, I recommend reviewing this proof and the proof of unique factroziation for the regular integers. I should also mention that one can modify the proof of unique factorization for $latex {\mathbb{Z}}$ as given in the course textbook as well (since it is a bit different than what we have done). Further, the course textbook does proof of unique factorization for $latex {\mathbb{Z}[i]}$ in Chapter 36, which is very similar to the proof sketched above (although the proof of Theorem~9 is very different.)

3. An application to $latex {y^2 = x^3 – 1}$.

We now consider the nonlinear Diophantine equation $latex {y^2 = x^3 – 1}$, where $latex {x,y}$ are in $latex {\mathbb{Z}}$. This is hard to solve over the integers, but by going up to $latex {\mathbb{Z}[i]}$, we can determine all solutions.

In $latex {\mathbb{Z}[i]}$, we can rewrite $$ y^2 + 1 = (y + i)(y – i) = x^3. \tag{1}$$
We claim that $latex {y+i}$ and $latex {y-i}$ are relatively prime. To see this, suppose that $latex {d}$ is a common divisor of $latex {y+i}$ and $latex {y-i}$. Then $latex {d \mid (y + i) – (y – i) = 2i}$. It happens to be that $latex {2i = (1 + i)^2}$, and that $latex {(1 + i)}$ is prime. To see this, we show the following.

Lemma 11 Suppose $latex {z}$ is a Gaussian integer, and $latex {N(z) = p}$ is a regular prime. Then $latex {z}$ is a Gaussian prime.

Proof: Suppose that $latex {z}$ factors nontrivially as $latex {z = ab}$. Then taking norms, $latex {N(z) = N(a)N(b)}$, and so we get a nontrivial factorization of $latex {N(z)}$. When $latex {N(z)}$ is a prime, then there are no nontrivial factorizations of $latex {N(z)}$, and so $latex {z}$ must have no nontrivial factorization. $latex \Box$

As $latex {N(1+i) = 2}$, which is a prime, we see that $latex {(1 + i)}$ is a Gaussian prime. So $latex {d \mid (1 + i)^2}$, which means that $latex {d}$ is either $latex {1, (1 + i)}$, or $latex {(1+i)^2}$ (up to multiplication by a Gaussian unit).

Suppose we are in the case of the latter two, so that $latex {(1+i) \mid d}$. Then as $latex {d \mid (y + i)}$, we know that $latex {(1 + i) \mid x^3}$. Taking norms, we have that $latex {2 \mid x^6}$.

By unique factorization in $latex {\mathbb{Z}}$, we know that $latex {2 \mid x}$. This means that $latex {4 \mid x^2}$, which allows us to conclude that $latex {x^3 \equiv 0 \pmod 4}$. Going back to the original equation $latex {y^2 + 1 = x^3}$, we see that $latex {y^2 + 1 \equiv 0 \pmod 4}$, which means that $latex {y^2 \equiv 3 \pmod 4}$. A quick check shows that $latex {y^2 \equiv 3 \pmod 4}$ has no solutions $latex {y}$ in $latex {\mathbb{Z}/4\mathbb{Z}}$.

So we rule out the case then $latex {(1 + i) \mid d}$, and we are left with $latex {d}$ being a unit. This es exactly the case that $latex {y+i}$ and $latex {y-i}$ are relatively prime.

Recall that $latex {(y+i)(y-i) = x^3}$. As $latex {y+i}$ and $latex {y-i}$ are relatively prime and their product is a cube, by unique factorization in $latex {\mathbb{Z}[i]}$ we know that $latex {y+i}$ and $latex {y-i}$ much each be Gaussian cubes. Then we can write $latex {y+i = (m + ni)^3}$ for some Gaussian integer $latex {m + ni}$. Expanding, we see that
\begin{equation}
y+i = m^3 – 3mn^2 + i(3m^2n – n^3).
\end{equation}
Equating real and imaginary parts, we have that
\begin{align}
y &= m(m^2 – 3n^2) \\
1 &= n(3m^2 – n^2).
\end{align}
This second line shows that $latex {n \mid 1}$. As $latex {n}$ is a regular integer, we see that $latex {n = 1}$ or $latex {-1}$.

If $latex {n = 1}$, then that line becomes $latex {1 = (3m^2 – 1)}$, or after rearranging $latex {2 = 3m^2}$. This has no solutions.

If $latex {n = -1}$, then that line becomes $latex {1 = -(3m^2 – 1)}$, or after rearranging $latex {0 = 3m^2}$. This has the solution $latex {m = 0}$, so that $latex {y+i = (-i)^3 = i}$, which means that $latex {y = 0}$. Then from $latex {y^2 + 1 = x^3}$, we see that $latex {x = 1}$.

And so the only solution is $latex {(x,y) = (1,0)}$, and there are no other solutions.

4. Other Rings

The Gaussian integers have many of the same properties as the regular integers, even though there are some differences. We could go further. For example, we might consider the following integer-like sets,
\begin{equation}
\mathbb{Z}(\sqrt{d}) = { a + b \sqrt{d} : a,b \in \mathbb{Z} }.
\end{equation}
One can add, subtract, and multiply these together in similar ways to how we can add, subtract, and multiply together integers, or Gaussian integers.

We might ask what properties these other integer-like sets have. For instance, do they have unique factorization?

More generally, there is a better name than ”integer-like set” for this sort of construction.

Suppose $latex {R}$ is a collection of elements, and it makes sense to add, subtract, and multiply these elements together. Further, we want addition and multiplication to behave similarly to how they behave for the regular integers. In particular, if $latex {r}$ and $latex {s}$ are elements in $latex {R}$, then we want $latex {r + s = s + r}$ to be in $latex {R}$; we want something that behaves like $latex {0}$ in the sense that $latex {r + 0 = r}$; for each $latex {r}$, want another element $latex {-r}$ so that $latex {r + (-r) = 0}$; we want $latex {r \cdot s = s \cdot r}$; we want something that behaves like $latex {1}$ in the sense that $latex {r \cdot 1 = r}$ for all $latex {r \neq 0}$; and we want $latex {r(s_1 + s_2) = r s_1 + r s_2}$. Such a collection is called a ring. (More completely, this is called a commutative unital ring, but that’s not important.)

It is not important that you explicitly remember exactly what the definition of a ring is. The idea is that there is a name for things that are ”integer-like” and that we might wonder what properties we have been thinking of as properties of the integers are actually properties of rings.

As a total aside: there are very many more rings too, things that look much more different than the integers. This is one of the fundamental questions that leads to the area of mathematics called Abstract Algebra. With an understanding of abstract algebra, one could then focus on these general number theoretic problems in an area of math called Algebraic Number Theory.

5. The rings $latex {\mathbb{Z}[\sqrt{d}]}$

We can describe some of the specific properties of $latex {\mathbb{Z}[\sqrt{d}]}$, and suggest how some of the ideas we’ve been considering do (or don’t) generalize. For a general element $latex {n = a + b \sqrt{d}}$, we can define the conjugate $latex {\overline{n} = a – b\sqrt {d}}$ and the norm $latex {N(n) = n \cdot \overline{n} = a^2 – d b^2}$. We call those elements $latex {u}$ with $latex {N(u) = 1}$ the units in $latex {\mathbb{Z}[\sqrt{d}]}$.

Some of the definitions we’ve been using turn out to not generalize so easily, or in quite the ways we expect. If $latex {n}$ doesn’t have a nontrivial factoriation (meaning that we cannot write $latex {n = ab}$ with $latex {N(a), N(b) \neq 1}$), then we call $latex {n}$ an irreducible. In the cases of $latex {\mathbb{Z}}$ and $latex {\mathbb{Z}[i]}$, we would have called these elements prime.

In general, we call a number $latex {p}$ in $latex {\mathbb{Z}{\sqrt{d}}}$ a prime if $latex {p}$ has the property that $latex {p \mid ab}$ means that $latex {p \mid a}$ or $latex {p \mid b}$. Of course, in the cases of $latex {\mathbb{Z}}$ and $latex {\mathbb{Z}[i]}$, we showed that irreducibles are primes. But it turns out that this is not usually the case.

Let us look at $latex {\mathbb{Z}{\sqrt{-5}}}$ for a moment. In particular, we can write $latex {6}$ in two ways as
\begin{equation}
6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 – \sqrt{-5}).
\end{equation}
Although it’s a bit challenging to show, these are the only two fundamentally different factorizations of $latex {6}$ in $latex {\mathbb{Z}[\sqrt{-5}]}$. One can show (it’s not very hard, but it’s not particularly illuminating to do here) that neither $latex {2}$ or $latex {3}$ divides $latex {(1 + \sqrt{-5})}$ or $latex {(1 – \sqrt{-5})}$ (and vice versa), which means that none of these four numbers are primes in our more general definition. One can also show that all four numbers are irreducible.

What does this mean? This means that $latex {6}$ can be factored into irreducibles in fundamentally different ways, and that $latex {\mathbb{Z}[\sqrt{-5}]}$ does not have unique factorization.

It’s a good thought exercise to think about what is really different between $latex {\mathbb{Z}[\sqrt{-5}]}$ and $latex {\mathbb{Z}}$. At the beginning of this course, it seemed extremely obvious that $latex {\mathbb{Z}}$ had unique factorization. But in hindsight, is it really so obvious?

Understanding when there is and is not unique factorization in $latex {\mathbb{Z}[\sqrt{d}]}$ is something that people are still trying to understand today. The fact is that we don’t know! In particular, we really don’t know very much when $latex {d}$ is positive.

One reason why can be seen in $latex {\mathbb{Z}[\sqrt{2}]}$. If $latex {n = a + b \sqrt{2}}$, then $latex {N(n) = a^2 – 2 b^2}$. A very basic question that we can ask is what are the units? That is, which $latex {n}$ have $latex {N(n) = 1}$?

Here, that means trying to solve the equation $$ a^2 – 2 b^2 = 1. \tag{2}$$
We have seen this equation a few times before. On the second homework assignment, I asked you to show that there were infinitely many solutions to this equation by finding lines and intersecting them with hyperbolas. We began to investigate this Diophantine equation because each solution leads to another square-triangular number.

So there are infinitely many units in $latex {\mathbb{Z}[\sqrt{2}]}$. This is strange! For instance, $latex {3 + 2 \sqrt{2}}$ is a unit, which means that it behaves just like $latex {\pm 1}$ in $latex {\mathbb{Z}}$, or like $latex {\pm 1, \pm i}$ in $latex {\mathbb{Z}[i]}$. Very often, the statements we’ve been looking at and proving are true ”up to multiplication by units.” Since there are infinitely many in $latex {\mathbb{Z}[\sqrt 2]}$, it can mean that it’s annoying to determine even if two numbers are actually the same up to multiplication by units.

As you look further, there are many more strange and interesting behaviours. It is really interesting to see what properties are very general, and what properties vary a lot. It is also interesting to see the different ways in which properties we’re used to, like unique factorization, can fail.

For instance, we have seen that $latex {\mathbb{Z}[\sqrt -5]}$ does not have unique factorization. We showed this by seeing that $latex {6}$ factors in two fundamentally different ways. In fact, some numbers in $latex {\mathbb{Z}[\sqrt -5]}$ do factor uniquely, and others do not. But if one does not, then it factors in at most two fundamentally different ways.

In other rings, you can have numbers which factor in more fundamentally different ways. The actual behaviour here is also really poorly understood, and there are mathematicians who are actively pursuing these topics.

It’s a very large playground out there.

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Math 420: Second Week Homework

Firstly, we have three administrative notes.

  1. I’ve posted the second homework set. You can find it here.
  2. I’ve also written solutions to the first homework set. You can find those here.
  3. After feedback from the first week, I’m setting stable office hours. My office hours will be from 1-3pm on Monday and 2:30-3:30pm on Tuesday (immediately following our class). [Or we can set up an appointment].

I’ll see you on Tuesday, when we will continue to talk about the Euclidean Algorithm and greatest common divisors.

 

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Math 420: First Week Homework and References

Firstly, there are three administrative notes.

  1. I’ve posted the first homework set. This is due on Thursday, and you can find it here.
  2. I haven’t set official office hour times yet. But I will have office hours on Monday from noon to 2pm on Monday, 1 Feb 2016, in my office at the Science Library.
  3. If you haven’t yet, I encourage you to read the syllabus.

We mentioned several good and interesting “number theoretic” problems in class today. I’d like to remind you of some of them, and link you to some additional places for information.

Pythagorean Theorem

We’ve found all primitive Pythagorean triples in integers, which is a very nice theorem for an hour. But I also mentioned some of the history of the Pythagorean Theorem and the significance of numbers and number theory to the Greeks.

I told the class a story about how the Pythagorean student who revealed that there were irrational numbers was stoned. This is apocryphal. In fact, there is little exact record, but his name was Hippasus and it is more likely that he was drowned for releasing this information.

For this and other reasons, the Pythagorean school of thought split into two sects, one from Pythagoras and one from Hippasus.

Goldbach’s Conjecture

Is it the case that every even integer is the sum of two primes? We think so. But we do not know.

I mentioned the Ternary Goldbach Conjecture, also known as the Weak Goldbach Conjecture, which says that every odd integer greater than $5$ is the sum of three odd primes. This was proved very recently. If you’re interested in what a mathematical paper looks like, you can give this paper a look. [Do not expect to be able to understand the paper — but it is interesting what sorts of tools can be used towards number theory]

Fermat’s Last Theorem

Are there nontrivial integer solutions to $X^n + Y^n = Z^n$ where $n \geq 3$?

This is one of the most storied and studied problems in mathematics. I think this has to do with how simple the statement looks. Further, we managed to fully classify all solutions when $n = 2$ in one class period. It doesn’t seem like it should be too hard to extend that to other exponents, does it?

If time and interest permits, we will return to this topic at the end of the course. There is no way that we could present a proof, or even fully motivate the proof. But we might be able to say a few words about how progress towards the theorem spurred and created mathematics, and maybe we can give a hint of the breadth of the ideas used to finally produce a proof.

Twin Prime Conjecture

Are there infinitely many primes $p$ such that $p+2$ is also prime? We think so, but we don’t know. Two years ago, we had absolutely no idea at all. Then Yitang Zhang had a brilliant idea (and not much later a graduate student named James Maynard had another brilliant idea) which allowed some sort of progress.

This culminated with the Polymath8 Project Bounded Gaps Between Primes. Math can be a social sport, and the polymath projects are massively collaborative online and open projects towards math problems. They’re still a bit new, and a bit experimental. But Polymath8 is certainly extremely successful.

What is known is that there exists at least one even number $H \leq 246$ such that $p$ and $p + H$ is prime infinitely often. In fact, James Maynard showed that you can make more complicated ensembles of prime distances.

The ideas that led to this result can likely be sharpened to give better results, but actually proving that there are infinitely many twin primes is almost certainly going to require a brand new idea and methodology.

The best related result comes from Chinese mathematician Chen Jingrun, who proved that every sufficiently large even integer can be written either as a sum of two primes, or as a sum of a prime and a number with exactly two prime factors. Although this seems very close, it is also likely that this idea cannot be sharpened further.

Writing Numbers as Sums of Squares, Cubes, and So On

Can every integer be written as the sum of three squares? What about four squares? More generally, is there a number $n$ so that every integer can be written as a sum of at most $n$ squares?

Similarly, is there a number $n$ so that every integer can be written as a sum of at most $n$ cubes? What about fourth powers?

These problems are all associated to something called Waring’s Problem, about which much is known and much is unknown.

We also asked which primes can be written as a sum of two squares. Although we might have a hard time finding those primes right now, you might try to show that if $p$ is a prime that can be written as a sum of two squares, then either $p$ is $2$, or $p = 4z + 1$ for some integer $z$. The reasoning is very similar to some of the reasoning done in class today.

Max’s Conjecture

For primitive Pythagorean triples $(a,b,c)$ with $a^2 + b^2 = c^2$, we showed that we can restrict out attention to cases where $a$ is odd, $b$ is even, and $c$ is odd. Max conjectured that those $c$ on the right are always of the form $4k + 1$ for some $k$, or equivalently $c$ is always an integer that leaves remainder $1$ after being divided by $4$.

We didn’t return to this in class, but we can now. First, note that since $c$ is odd, we can write $c$ as $2z + 1$ for some $z$. But we can do more. We can actually write $c$ as either $4z + 1$ or $4z + 3$. (Can you prove this?)

Max conjectured that it is always the case that $c = 4z + 1$. So we might ask, “What if $c = 4z + 3$?”

Writing $a = 2x + 1$ and $b = 2y$, we get the equation

$$ \begin{align}
a^2 + b^2 &= c^2 \\
4x^2 + 4x + 1 + 4y^2 &= 16z^2 + 24z + 3, \end{align}$$

which can be rewritten as
$$ 4x^2 + 4x + 4y^2 = 16z^2 + 24z + 2.$$
You can divide by $2$. Then we ask: what’s the problem? Why is this bad? (It is, and it’s very similar to some questions we asked in class.)

So Max’s Conjecture is true, and every number appearing as $c$ in a primitive Pythagorean triple is of the form $c = 4z + 1$ for some integer $z$.

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