The Oldest Unsolved Problem in Math ?

AnthonyRBrown · 2024-03-23 10:01:00

Now this is interesting,I only just discovered this...

https://www.youtube.com/watch?v=Zrv1EDIqHkY

The four perfect numbers below were the only ones found for over a thousand years!
What is interesting is that there are a lot of patterns related to each of them,so it should be easy to find the next perfect number?

6
28
496
8128

The person talking in the video above said the pattern that really blows his mind,is if you write the numbers in binary you get...

6 = 110
28 = 11100
496 = 111110000
8128 = 1111111000000

The count of the 1's on the left are 2,3,5,7 and the 0's are 1,2,4,6

So the next perfect number should be easy to find,if we forget about the start!!! the 2,and the 1 are the beginning of a cycle is the way I look at it

All the next perfect numbers should continue like,the 1's are +2 and the 0's are +2

Which makes the next perfect number in binary = 11111111100000000 ?

Using the Binary to Decimal converter below...

https://www.rapidtables.com/convert/number/binary-to-decimal.html

We get the next perfect number as = 130816 ?

Is the above true? and can we continue with all the next perfect numbers!

A.R.B

Bob · 2024-03-23 22:08:58

Wow! I thought you had stumbled onto something here. But I did a bit of googling and found this

https://byjus.com/maths/perfect-numbers … 033550336.

If you look at how they are formed from that formula you'll see the connection with binary numbers. You just hadn't quite got the general term. Nice try

Bob

AnthonyRBrown · 2024-03-23 22:54:10

Bob wrote:

Wow! I thought you had stumbled onto something here. But I did a bit of googling and found this
https://byjus.com/maths/perfect-numbers … 033550336.
If you look at how they are formed from that formula you'll see the connection with binary numbers. You just hadn't quite got the general term. Nice try
Bob

Ok! So the next 5th perfect number is not 130816 ? it is 33550336 ?

If we convert that to Binary we get 1111111111111000000000000 ?

Let's compare that with the first 4 perfect numbers below...

6 = 110
28 = 11100
496 = 111110000
8128 = 1111111000000
33550336 = 1111111111111000000000000

What happened to the (All the next perfect numbers should continue like,the 1's are +2 and the 0's are +2) ?

The next 5th perfect number should be...

130816 = 11111111100000000

And it's not! Oh dear this looks like a very hard problem! I will have to give this more than 5 minutes!

A.R.B

Bob · 2024-03-24 01:47:01

Good luck. It's been around a while and Euler spent a lot of time on it with some success.

Euclid's formula is

where the second term is prime.

In binary all powers of 2 are 1 followed by some zeros. For example 2^5 = 100000 If you subtract 1 the result, in binary is a string of 1s eg. 31 is 11111 in binary.

So when those terms are multipled together you get the string of 1s from the second term and a string of zeros after that because of the powers of 2.

So the format is always

But the question is how many 1s and how many 0s ?

Bob

Bob · 2024-03-24 03:59:17

Proof of Euclid's formula.

It is important to understand that this proof is of the form A => B. This does not mean that B => A

In other words, a number with this format can be shown to be perfect. But a number that is perfect may not have this format ie. there may be other perfect numbers that are not generated by this format. So far no mathematician has succeeded in finding such a number or proving the format produces the only perfect numbers.

Background for those who haven't met this before.

A number such a 6 is said to be perfect because its divisors (other than 6 itself) add uo to 6 (= 1+2+3)

Another such is 28 (= 1+2+4+7+14)

Euclid explored this and found a formula for generating perfect numbers.

If a number has the form

where (2^{p} - 1) is prime, will always be perfect.

6 = 2 x 3 has this format. 28 = 2^2 x 7 has too.
This table with p from 2 to 6, shows this formula in action

Let's see why this happens for some of these and why it doesn't when the second term is not prime.

Without knowing what they are, we can say that 6 has 4 factors. Firstly there's the powers of 2 (2^0 and 2^1) and then the factors formed by multiplying the other term (3) by each power of 2 ( 2^0 times 3 and 2^1 times 3). This last factor is not counted for perfect number purposes. The factors we do count are 1, 2 and 3 and these add to 6.

28 = 2^2 times 7. From this we can say that 28 has 6 factors. Firstly the powers of 2 (2^0, 2^1 and 2^2) and then the factors formed by multiplying the other term (7) by each power of 2 (2^0 times 7, 2^1 times 7 and 2^2 times 7). Again the last of these isn't counted.

With numbers, this is what my proof does.

Add up

This is the formula with p=2.

Another example for the factors of 28

Next what happens when, say, p = 4. 2^p -1 is 15 which is not prime.

When the second term is prime there are always 2 times p factors and the highest is disgarded. When the second term isn't prime there are a lot more factors: the powers of two; the powers of two multiplied by (in this case) 3; the powers of two multiplied by 5; the powers of two multiplied by 15. Way too many factors to ever stand a chance of being perfect. Only second term prime keeps the total factors down to 2p altogether.

Proof

Consider a number with this format. You can make up factors in two ways (1) by multiplying the prime part by any of the powers of two from 2^0 up to 2^(p-2) and (2) by listing all the powers of two from 1 up to 2^(p-1). I'll add these two sets up.

Note: that first part stops at 2^(p-2) because the next term would generate the number itself and this is excluded when considering factors of the number.

Most of these powers of two cancel. Also the bracketed sum is 2^(p-1) less one so

which is the number I started with.

Bob

AnthonyRBrown · 2024-03-24 07:03:17

Great stuff Bob iit looks like you have been there before

For the moment it looks like there is enough of a problem finding something in the first 5 numbers! the fifth number is a real Monster and looks like it's a made up hoax,I have never seen such a difference within a small group of related numbers?

Bob · 2024-03-24 22:23:40

Having come up with a proof I was very keen to get it typed up for a post. Looking it over I decided it may be a bit 'heavy' for some members. Usually when I find the algebra is getting tough I ease the student in carefully with some number cases first. With this in mind I have edited my previous post by adding a table of formula values from 2 to 6. This shows that only second term prime gives a perfect number. Then I put numbers into my 'proof' to show it working for 6 and 28 and explained why it fails when p = 4 because 15 isn't prime.

Also I think it's easier to follow the proof if I start with the factors which are powers of two and then follow with the ones made by multiplying the prime by the powers of two. So I have switched these bits around in the proof.

In binary 2^p - 1 will always be a string of 1s and 2^(p-1) will always be a 1 followed by a string of zeros. So that would account for your binary observations. But, to get perfects, you need the string of 1s to be prime. There's no formula in decimals for generating primes so there cannot be one in binary either. So you're still stuck deciding which string of 1s is going to produce a prime.

Bob

AnthonyRBrown · 2024-03-26 22:22:19

Below are the first 10 Perfect numbers,so if we can find a related Pattern amongst them? then I might be able to pull of a another FLT Demonstration only this time is will be a PND Perfect Number Demonstration!

6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
2658455991569831744654692615953842176
191561942608236107294793378084303638130997321548169216

How to Find Perfect Number?

We can easily find perfect numbers using the following steps:

Perfect numbers are positive integers n such that n=s(n),

where, s(n) is the restricted divisor function (i.e., the sum of proper divisors of n), or equivalently

sigma(n)=2n,

where sigma(n) is the divisor function (i.e., the sum of divisors of n including n itself). For example, the first few perfect numbers are 6, 28, 496, 8128, … (OEIS A000396), since

6 = 1+2+3

28 = 1+2+4+7+14

496 = 1+2+4+8+16+31+62+124+248,

Perfect Numbers Solved Examples

Solved Example: Find out if 496 is a perfect number.

Solution: The proper factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, and 248

1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496

496 is perfect.

Math Is Fun Forum

#1 2024-03-23 10:01:00

The Oldest Unsolved Problem in Math ?

#2 2024-03-23 22:08:58

Re: The Oldest Unsolved Problem in Math ?

#3 2024-03-23 22:54:10

Re: The Oldest Unsolved Problem in Math ?

#4 2024-03-24 01:47:01

Re: The Oldest Unsolved Problem in Math ?

#5 2024-03-24 03:59:17

Re: The Oldest Unsolved Problem in Math ?

#6 2024-03-24 07:03:17

Re: The Oldest Unsolved Problem in Math ?

#7 2024-03-24 22:23:40

Re: The Oldest Unsolved Problem in Math ?

#8 2024-03-26 22:22:19

Re: The Oldest Unsolved Problem in Math ?

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