Prove that if $2^n-1$ is a Mersenne prime number, then
\[N=2^{n-1}(2^n-1)\] is a perfect number.

On the other hand, prove that every even perfect number $N$ can be written as $N=2^{n-1}(2^n-1)$ for some Mersenne prime number $2^n-1$.

Definitions.

In this post, a divisor of a positive integer means a positive divisor.

The Sum of Divisors Function $\sigma(n)$

For a positive integer $n$, we denote by $\sigma(n)$ the sum of all divisors of $n$.
(We call $\sigma$ the sum of the divisors function.)

For example, consider the number $12$.
The divisors of $12$ are $1, 2, 3, 4, 6, 12$.

Thus, the sum of the divisors of $12$ is
\[\sigma(12)=1+2+3+4+6+12=28.\]

Perfect Numbers

A positive integer $n$ is called a perfect number if the sum of all proper divisors are $n$ itself.
Here, a proper divisor of $n$ means a divisor of $n$ that is not $n$ itself.

Alternatively and equivalently, we can define a perfect number to be a positive integer $n$ such that the sum of all divisors of $n$ (including $n$ itself) is $2n$.

In terms of the function $\sigma(n)$, an integer $n$ is a perfect number if $\sigma(n)=2n$.

Examples of Perfect Numbers

For example, the number $6$ is the first perfect number.

In fact, the proper divisors of $6$ are $1, 2, 3$ and their sum is $1+2+3=6$.
We also can use the second equivalent definition to check that $6$ is a perfect number.
We have
\begin{align*}
\sigma(6)=1+2+3+6=12=2\cdot 6.
\end{align*}

The next perfect number is $28$.
To see this, we find the divisors of $28$ and sum them up, and we have
\[\sigma(28)=1+2+4+7+14+28=56=2\cdot 28.\]

(Wait, did you notice that this problem is posted on June 28th?
I propose to call June 28th Perfect Number Day or simply Perfect Day.)

Can you find the next perfect number?

I give you a hint: Look at the problem number of this post.

Yes, the third perfect number is

Unsolved Problems in Number Theory

All the perfect numbers people have found so far are all even numbers.

According to Wikipedia, 49 Mersenne primes are known as of January 2016.

Even though we don’t know whether there are infinitely even perfect numbers, we do know how they should look like thanks to the great mathematician Euclid and Euler.

This is the problem of this post.

Mersenne Prime Numbers

A number of the form $2^n-1$ is called a Mersenne number.
The first few Mersenne numbers are $1, 3, 7, 15, 31$, corresponding to $n=1, 2, 3, 4, 5$.

As you can see from these examples, not all Mersenne numbers are prime numbers.
If a Mersenne number is a prime number, then we call it a Mersenne prime number.

Thus $2^2-1=3, 2^3-1=7, 2^5-1=31$ are examples of Mersenne prime numbers, but $2^4-1=15=3\cdot 5$ is not a Mersenne prime number.

The problem claims that:

Thus, even perfect numbers and Mersenne prime numbers are one-to-one correspondence: If you find one, then you find the other.

As we said earlier, we don’t know whether there are infinitely many perfect numbers.
Thus, we also don’t know whether there are infinitely many Mersenne prime numbers.

Properties of the Sum of Divisors Function $\sigma$

To prove the problem, let us first study the sum of divisors function $\sigma(n)$.

Example How to Use the Formula

Before the proof of Theorem, let us give an example to illustrate how to use the formula in Theorem.
Consider $24$.
The prime factorization is $24=2^3\cdot 3$.
Thus, by Theorem the sum of the divisors of $24$ is
\begin{align*}
\sigma(24)&=\frac{2^{3+1}-1}{2-1}\cdot \frac{3^{1+1}-1}{3-1}\\
&=15\cdot 4 =60.
\end{align*}

Just to make sure, let us list divisors of $24$:

\[1, 2, 3, 4, 6, 8, 12, 24\]

and it is checked that the sum of these divisors is $60$ and it (of course) coincides with the result we obtained using Theorem.

Proof of Theorem

From the prime factorization $n=p_1^{a_1}\dots p_k^{a_k}$, every divisor of $n$ is of the form
\[n=p_1^{b_1}\dots p_k^{b_k},\] where $0 \leq b_i \leq a_i$ for $i=1, \dots, k$.

If the product
\[(1+p_1+p_1^2+\cdots+p_1^{a_1})\cdots (1+p_k+p_k^2+\cdots+p_k^{a_k})\] is expanded, then this becomes some of
\[n=p_1^{b_1}\dots p_k^{b_k},\] where each $b_i$ runs from $0$ to $a_k$.
Thus, it is the sum of the divisors of $n$.
This proves the first equality in the formula.

By the formula for a geometric series, we have
\begin{align*}
1+p_i+p_i^2+\cdots+p_i^{a_i}=\frac{p^{a_i+1}-1}{p_i-1}.
\end{align*}
This yields the second equality.
This completes the proof of Theorem.

As a corollary of this theorem, we see that the sum of the divisors function $\sigma(n)$ is multiplicative.

Proof of Corollary

Let
\[m=p_1^{a_1}\dots p_k^{a_k} \text{ and } n=q_1^{b_1}\dots q_l^{b_l}\] be the prime factorizations of $m$ and $n$.
Since $m$ and $n$ are relatively prime, each pair of $p_i$ and $q_j$ is distinct.
Thus,
\[nm=p_1^{a_1}\dots p_k^{a_k}q_1^{b_1}\dots q_l^{b_l}\] is the prime factorization of $mn$.

By the theorem above, we have
\begin{align*}
\sigma(m)=\prod_{i=1}^k \frac{p^{a_i+1}-1}{p_i-1}\\[6pt] \sigma(n)=\prod_{i=1}^l \frac{q^{b_i+1}-1}{q_i-1}
\end{align*}
and thus we have
\begin{align*}
\sigma(m)\sigma(n)=\prod_{i=1}^k \frac{p^{a_i+1}-1}{p_i-1}\cdot \prod_{i=1}^l \frac{q^{b_i+1}-1}{q_i-1}=\sigma(mn),
\end{align*}
where the second equality also follows from the theorem above.
This completes the proof of Corollary.

Proof of the Problem about Perfect Numbers and Mersenne Prime Numbers

If $2^n-1$ is a Mersenne Prime, then $2^{n-1}(2^n-1)$ is a Perfect Number

Suppose that $p:=2^n-1$ is a Mersenne prime number.
Then the sum of the divisors of $N=2^{n-1}p$ is
\begin{align*}
\sigma(N)&=\sigma(2^{n-1}p)\\
&=(1+2+2^2+\cdots+2^{n-2}+2^{n-1})(1+p) && \text{by Theorem}\\
&=(2^n-1)2^n\\
&=2\cdot 2^{n-1}(2^n-1)\\
&=2N.
\end{align*}

Therefore, the number $N$ is a perfect number.

If $N$ is an even perfect number, then $N=2^{n-1}(2^{n}-1)$ with Mersenne Prime $2^{n}-1$.

Suppose now that $N$ is an even perfect number.
Since $N$ is even, we can write
\[N=2^am\] for some positive integers $a$ and an odd integer $m$.

Since $N$ is a perfect number, we have
\[\sigma(N)=2N=2^{a+1}m.\] We compute
\begin{align*}
\sigma(N)&=\sigma(2^a)\sigma(m) &&\text{by Corollary}\\
&=(2^{a+1}-1)\sigma(m) &&\text{by Theorem}.
\end{align*}

Hence we have
\[2^{a+1}m=(2^{a+1}-1)\sigma(m).\]

Solving this to $\sigma(m)$, it yields that
\begin{align*}
\sigma(m)&=\frac{2^{a+1}m}{2^{a+1}-1}\\[6pt] &=\frac{(2^{a+1}-1+1)m}{2^{a+1}-1}\\[6pt] &=m+\frac{m}{2^{a+1}-1}.
\end{align*}

As both $\sigma(m)$ and $m$ are integers, so is $\frac{m}{2^{a+1}-1}$.
Since $a > 0$, the denominator $2^{a+1}-1 > 1$.
Hence $\frac{m}{2^{a+1}-1}$ is a proper divisor of $m$.

Now observe the equality we obtained above again:
\[\sigma(m)=m+\frac{m}{2^{a+1}-1}.\] The left hand side $\sigma(m)$ is the sum of all the divisors of $m$.
The right hand side is the sum of two divisors of $m$: $m$ itself and $\frac{m}{2^{a+1}-1}$.

It follows that $m$ has only two divisors, and we must have
\[1=\frac{m}{2^{a+1}-1},\] or equivalently
\[m=2^{a+1}-1.\]

The only integers that have exactly two divisors are prime numbers.
Thus, $m$ is a prime number, and it is of the form $m=2^{a+1}-1$.
Therefore $m$ is a Mersenne prime number, and the even perfect number $N$ can be written as
\[N=2^am=2^a(2^{a+1}-1)\] with Mersenne prime $m=2^{a+1}-1$ as required.

This completes the proof of the problem.

The Largest Perfect Number

The largest Perfect Number we know is

with 44,677,235 digits.

By Problem, we know that

is the largest Mersenne prime number we know.

(Reference: Wikipedia)

Comments

I’m happy that I could post the 496th article about perfect numbers on the Perfect Day June 28th, 2017.

Is 2017 a perfect number?

No!! But 2017 is a prime number!!

The post Even Perfect Numbers and Mersenne Prime Numbers 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.

(a) Show that if a group $G$ has the following order, then it is not simple.

1. $28$
2. $496$
3. $8128$

(b) Show that if the order of a group $G$ is equal to an even perfect number then the group is not simple.

Hint.

Use Sylow’s theorem.
(See the post Sylow’s Theorem (summary) to review the theorem.)

Proof.

(a) A group of the following order is not simple

(1) A group of order $28$

Note that $28=2^2\cdot 7$. The number $n_7$ of the Sylow $7$-subgroups of $G$ satisfies
\[n_7 \equiv 1 \pmod{7} \text{ and } n_7|2^2.\] Thus, the only possible value is $n_7=1$. The unique Sylow $7$-subgroup is a proper nontrivial normal subgroup of $G$, hence $G$ is not simple.

(2) A group of order $496$

Note that $496=2^4\cdot 31$. By the same argument as in (1), there is a normal Sylow $31$-subgroup in $G$, hence $G$ is not simple.

(3) A group of order $8128$

We have $8128=2^6\cdot 127$, where $127$ is a prime number.
Again the same reasoning proves that the group $G$ has the unique normal Sylow $127$-subgroup in $G$, hence $G$ is not simple.

(b) If the order is an even perfect number, a group is not simple

From elementary number theory, all even perfect numbers are of the form
\[2^{p-1}(2^p-1),\] where $p$ is a prime number and $2^p-1$ is also a prime number.
(For a proof, see the post “Even Perfect Numbers and Mersenne Prime Numbers“.)

Suppose the order of a group $G$ is $2^{p-1}(2^p-1)$, with prime $p$, $2^p-1$.
Then the number $n_{2^p-1}$ of Sylow $(2^p-1)$-subgroup satisfies
\[n_{2^p-1}\equiv 1 \pmod {2^p-1} \text{ and } n_{2^p-1}|2^{p-1}.\] These force that $n_{2^p-1}=1$.
Therefore the group $G$ contains the unique normal Sylow $(2^p-1)$-subgroup, hence $G$ is not simple.

Similar problem

For an analogous problem, check out: Groups of order 100, 200. Is it simple?

Comment.

In about 300 BC Euclid showed that a number of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime numbers.
(If $2^p-1$ is a prime number, then $p$ must be a prime.)
The converse was proved by Euler in the 18th century. Namely, Euler proved that any even perfect number is of the form $2^{p-1}(2^p-1)$ with prime $2^p-1$.

The number of the form $2^p-1$ is called a Mersenne number and if it is a prime number, then it is called a Mersenne Prime.
It is unknown whether there are infinitely many Mersenne prime numbers.

Also it is unknown whether there is an odd perfect number.

The post If the Order is an Even Perfect Number, then a Group is not Simple first appeared on Problems in Mathematics.