Prove that there are infinitely many prime numbers in ONE-LINE.

Background

There are several proofs of the fact that there are infinitely many prime numbers.
Proofs by Euclid and Euler are very popular.

In this post, I would like to introduce an elegant one-line proof published by Sam Northshield in 2015.

Because the published paper really contains only one line, it is hard to cite only a part of it.
So I decided to cite all of his proof word for word.

Proof by Sam Northshield (2015).

More details in this one-line proof.

The above proof was given by Sam Northshield in 2015.
Let us give more details hidden in this one line proof.

Suppose that there are only finitely many prime numbers $p_1, p_2, \dots, p_n$.
Since prime numbers must be greater than $1$, we have
\[\sin\left(\, \frac{\pi}{p_i} \,\right) > 0\] for any $i=1, \dots, n$.
Thus, the product
\[\prod_{i=1}^n\sin\left(\, \frac{\pi}{p_i} \,\right)\] is still positive since the product of finitely many positive numbers is positive.
This explains the first inequality in (*).

Recall the following basic property of the sine function.
For any integer $m$, we have
\[\sin(\theta+2\pi m)=\sin(\theta)\] for any $\theta \in \R$.

We have
\begin{align*}
\frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} &=\frac{\pi}{p_i}+\frac{2\pi \prod_{j=1}^n p_j}{p_i}.
\end{align*}
Note that since $\prod_{j=1}^n p_j$ is the product of all $p_1, \dots, p_n$, one factor is $p_i$.
Hence
\[m_i:=\frac{\prod_{j=1}^n p_j}{p_i}\] is just an integer, not a fraction.
It follows from this observation that we have for each $i=1, \dots, n$
\begin{align*}
\sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right)&=\sin\left(\, \frac{\pi}{p_i}+2\pi m_i \,\right)\\
&=\sin\left(\, \frac{\pi}{p_i} \,\right)
\end{align*}
by the property of the sine function mentioned above.

Taking the product of these over all $i=1, \dots, n$, we obtain
\begin{align*}
\prod_{i=1}^n \sin \left(\, \frac{\pi}{p_i} \,\right)=\prod_{i=1}^n \sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right).
\end{align*}
This is the first equality in (*).

To see the last equality in (*), we consider the number
\[1+2\prod_{j=1}^n p_j\] in the numerator.

If this is a prime number, then it must be one of $p_1, \dots, p_n$.
If it is not a prime number, then it is divisible by some prime number $p_1, \dots, p_n$.
In either case, there is a prime number $p_{i_0}$ among $p_1, \dots, p_n$ such that
\[\frac{1+2\prod_{j=1}^n p_j}{p_{i_0}}\] is an integer.
Therefore, we have
\[\sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_{i_0}} \,\right)=0.\]

Since one of the factors is zero, the product
\[\prod_{i=1}^n \sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right)\] is also zero.
This proves the last equality in (*).
The inequality obtained in (*) is clearly a contradiction.
Hence there must be infinitely many prime numbers.
This completes the proof.

Comment.

The one-line proof of Northshield is very clever and elegant.
But to explain the proof to high-school students, I feel that I need to probably decipher the proof and give more explanations as I did in this post.

Once you understand the details, the one-line proof is a very convenient and beautiful way to hide the details.

Reference

The one-line proof was published in the paper

Northshield, Sam. A one-line proof of the infinitude of primes.
Amer. Math. Monthly 122 (2015), no. 5, 466.

The post A One-Line Proof that there are Infinitely Many Prime Numbers first appeared on Problems in Mathematics.

Let $a, b$ be relatively prime integers and let $p$ be a prime number.
Suppose that we have
\[a^{2^n}+b^{2^n}\equiv 0 \pmod{p}\] for some positive integer $n$.

Then prove that $2^{n+1}$ divides $p-1$.

Proof.

Since $a$ and $b$ are relatively prime, at least one of them is relatively prime to $p$.
Without loss of generality let us assume that $b$ and $p$ are relatively prime.

Then the given equality becomes
\begin{align*}
a^{2^n}\equiv -b^{2^n} \pmod{p} \\
\iff \left( \frac{a}{b}\right)^{2^n} \equiv -1 \pmod{p}.
\end{align*}
Taking square of both sides we obtain
\[\left( \frac{a}{b}\right)^{2^{n+1}} \equiv 1 \pmod{p}.\]

Now, we can think of these congruences as equalities of elements in the multiplicative group $(\Z/p\Z)^{\times}$ of order $p-1$:
\[ \left( \frac{a}{b}\right)^{2^n} = -1 \text{ and } \left( \frac{a}{b}\right)^{2^{n+1}} =1 \text{ in } (\Z/p\Z)^{\times}.\]

Note that the second equality yields that the order of the element $a/b$ divides $2^{n+1}$.
On the other hand, the first equality implies that any smaller power of $2$ is not the order of $a/b$.
Thus, the order of the element $a/b$ is exactly $2^{n+1}$.

In general, the order of each element divides the order of the group.
(This is a consequence of Lagrange’s theorem.)

Since the order of the group $(\Z/p\Z)^{\times}$ is $p-1$, it follows that $2^{n+1}$ divides $p-1$.
This completes the proof.

The post Number Theoretical Problem Proved by Group Theory. $a^{2^n}+b^{2^n}\equiv 0 \pmod{p}$ Implies ^{n+1}|p-1$. first appeared on Problems in Mathematics.

Here is the list of mathematical facts about the number 2017 that you can brag about to your friends or family as a math geek.

2017 is a prime number

Of course, I start with the fact that the number 2017 is a prime number.

The previous prime year was 2011.
The next prime year is 2027 and it is actually a twin prime year (2027 and 2029 are both primes).

• 2017th prime number is 17539.
• Combined number 201717539 is also prime.
• Yet combined number 175392017 is composite.
• 2017 is 306th prime number. $306=2\cdot 3^2\cdot 17$ contains a prime factor 17.
• 2017+2+0+1+7=2027 is the next prime year.

You may find more prime years from the list of one million primes that I made.

2017 is not a Gaussian prime

The number 2017 is congruent to 1 mod 4. (When we divide 2017 by 4, the remainder is 1.)
Such a number can be factored in the ring of Gaussian integers $\Z[i]$, where $i=\sqrt{-1}$. Explicitly we have
\[2017=(44+9i)(44-9i).\]

2017 is not an Eisenstein prime

The number 2017 can be factored in the ring of Eisenstein integers $\Z[\omega]$, where $\omega=e^{2\pi i/3}$ is a primitive third root of unity, as
\[2017=(-7-48\omega^2)(41+48\omega^2).\]

2017 is a sum of squares

We can write 2017 as a sum of two squares:
\[2017=44^2+9^2.\]

2017 is a part of Pythagorean triple

A triple $(a, b, c)$ of integers is called a Pythagorean triple if we have
\[a^2+b^2=c^2.\] The triple
\[(1855, 792, 2017)\] is a Pythagorean triple because we have
\[1855^2+792^2=2017^2.\]

(To obtain these numbers note that in general for any integers $m>n>0$, the triple $(a, b, c)$, where
\[a=m^2-n^2, b=2mn, c=m^2+n^2\] is a Pythagorean triple by Euclid’s formula.
Since we know $2017=44^2+9^2$, apply this formula with $m=44, n=9$.)

A Pythagorean triple $(a, b, c)$ is said to be primitive if the integers $a, b, c$ are coprime. A Pythagorean triple obtained from Euclid’s formula is primitive if and only if $m$ and $n$ are coprime. In our case, $m=44$ and $n=9$ are coprime, the Pythagorean triple $(1855, 792, 2017)$ is primitive.

By the way, Carl Friedrich Gauss passed away on February 23rd 1855.
(Reference: Wikipedia Carl Friedrich Gauss.)

2017 is a sum of three cubes

The number 2017 can be expressed as a sum of three cubes of primes:
\[2017=7^3+7^3+11^3.\]

2017 appears in $\pi$

The number 2017 appear in the decimal expansion of $\pi=3.1415…$.
Look at the last four numbers of $\pi=3.1415…2017$ truncated to $8900$ decimal places.

Here is the proof.

The number 2017 does not appear in the decimal expansion of $2017^{2017}$.

Exam problem using 2017

Let
\[A=\begin{bmatrix}
-1 & 2 \\
0 & -1
\end{bmatrix} \text{ and } \mathbf{u}=\begin{bmatrix}
1\\
0
\end{bmatrix}.\] Compute $A^{2017}\mathbf{u}$.
This is one of the exam problems at the Ohio State University.
Check out the solutions of this problem here.

How many prime numbers are there?

2017 is a prime number. How many prime numbers exist?

In fact, there are infinitely many prime numbers.

Please check out the post

A One-Line Proof that there are Infinitely Many Prime Numbers.

As the title suggests, the proof is only in one-line.

More fun with 2017?

If you know or come up with more interesting properties of the number 2017, please let me know.

I hope 2017 will be a wonderful year for everyone!!

The post Mathematics About the Number 2017 first appeared on Problems in Mathematics.

Use Lagrange’s Theorem in the multiplicative group $(\Zmod{p})^{\times}$ to prove Fermat’s Little Theorem: if $p$ is a prime number then $a^p \equiv a \pmod p$ for all $a \in \Z$.

 

Before the proof, let us recall Lagrange’s Theorem.

Lagrange’s Theorem

If $G$ is a finite group and $H$ is a subgroup of $G$, then the order $|H|$ of $H$ divides the order $|G|$ of $G$.

Proof.

If $a=0$, then we clearly have $a^p \equiv a \pmod p$.
So we assume that $a\neq 0$.
Then $\bar{a}=a+p\Z \in (\Zmod{p})^{\times}$.

Let $H$ be a subgroup of $(\Zmod{p})^{\times}$ generated by $\bar{a}$.
Then the order of the subgroup $H$ is the order of the element $\bar{a}$.

By Lagrange’s Theorem, the order $|H|$ divides the order of the group $(\Zmod{p})^{\times}$, which is $p-1$.
So we write $p-1=|H|m$ for some $m \in \Z$.

Therefore, we have
\begin{align*}
\bar{a}^{p-1}=\bar{a}^{|H|m}=\bar1^m=\bar1.
\end{align*}
(Note that this is a computation in $(\Zmod{p})^{\times}$.)

This implies that we have
\[a^{p-1}\equiv 1 \pmod p.\] Multiplying by $a$, we obtain
\[a^{p}\equiv a\pmod p,\] and hence Fermat’s Little Theorem is proved.

The post Use Lagrange's Theorem to Prove Fermat's Little Theorem first appeared on Problems in Mathematics.

Find the largest prime number less than one million.

What is a prime number?

A natural number is called a “prime number” if it is only divisible by $1$ and itself.
For example, $2, 3, 5, 7$ are prime numbers, although the numbers $4,6,9$ are not.

The prime numbers have always fascinated mathematicians.
There are a lot of unsolved problems related to prime numbers.

There are many special types of prime numbers named after famous mathematicians.
My favorites are Mersenne primes, Fermat primes, and Wagstaff primes.

• A natural number of the form
  \[2^n-1\] is called a Mersenne number.
• A Mersenne prime is a prime number of the form
  \[2^p-1.\]
• A natural number of the form
  \[2^{2^n}+1 \] is called a Fermat number.
• A Fermat prime is a prime number of the form
  \[2^{2^n}+1.\]
• A Wagstaff prime is a prime number of the form
  \[\frac{2^p+1}{3}.\]

Unsolved problems

For these prime numbers the followings are still unknown.

• Are there infinitely many Mersenne/Fermat/Wagstaff prime numbers?
• Are there infinitely many nonprime Fermat numbers?
• Are there infinitely many composite Mersenne number $2^p-1$ for a prime $p$?

What is the largest prime number less than one million.

It is known for a long time (Euclid’s Elements (circa 300 BC)) that there are infinitely many primes.

Here are the first $95$ prime numbers. These are all prime numbers less than $500$.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,
67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137,
139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211,
223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283,
293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379,
383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461,
463, 467, 479, 487, 491, 499.

List of prime numbers less than one million.

In fact, there are $78,498$ prime numbers less than $1,000,000$=one million.
To list them here takes a lot of space, so I created a PDF file of the list of primes less than one million.

It takes $95$ pages just to list $78498$ prime numbers less than one million.

Prime numbers less than one million

From this list, we see that

the largest prime numbers less than one million is $999983$.
(The last number in the PDF file.)

Other Facts

Here are several facts that we can find from the list (with time and energy)

• The largest twin prime pair less than one million is $999959$ and $999961$.
• The second largest twin prime pair less than one million is $999611$ and $999613$.
• The third largest twin prime pair less than one million is $999431$ and $999433 $.
• There are 7 Mersenne primes less than one million. These Mersenne primes are
  \[3, 7, 31, 127,8191, 131071, 524287.\]
• The know Fermat prime numbers are all less than one million.These are
  \[ 3, 5, 17, 257, 65537.\]
• $11$ is the only prime number containing only the decimal digit 1 and less than one million. (The second largest such prime is $1111111111111111111$.)
• Wagstaff prime numbers less than one million are \[3, 11, 43, 683, 2731, 43691, 174763.\]

Try to find an interesting property of prime numbers from the list of primes <100000.

The post Find the Largest Prime Number Less than One Million. first appeared on Problems in Mathematics.