### Euler 625 Details

Verfasst:

**29.04.2010 12:50**I'm the author of the original project about computing (6,2,5): http://euler.free.fr/

The project is running since 11 years, but I think the new yoyo project will outperform our project in a few days.

The goal is to compute solutions to the equation:

a^6 + b^6 = c^6 + d^6 + e^6 + f^6 + g^6

We use the notation (6,2,5) to express the fact that the equation is at 6th power, has 2 left terms and 5 right terms.

Reminder: ^ means power. Thus a^6 = a*a*a*a*a*a.

For example, 40^6 = 4096000000, as you can see, a^6 grows very quickly.

Why is this equation particularly interesting ?

In fact, it is not very useful.

We hope to find a solution where one of the terms a,b,c,d,e,f or g is zero, in other words, this means that we are searching (6,1,5) or (6,2,4).

Why are we searching for (6,1,5) or (6,2,4) ?

Some mathematicians conjectured that we could find solutions of equations (k,m,n), where k=m+n.

In other words, we can find a combination of 6 terms at the 6th power that lead to 0 when added or subtracted.

Currently, we found solutions for the following equations:

(4,1,3), (4,2,2)

(5,1,4) (5,2,3)

(6,3,3)

(8,3,5) (8,4,4)

For example, for (6,3,3), Subba-Rao found the following result in 1934:

23^6+15^6+10^6=22^6+19^6+3^6

in fact, there is an infinite number of solutions for (6,3,3)

To extend the above list of results, the easier solutions to reach are, in increasing difficulty order:

(6,2,4), (7,3,4), (7,2,5), (7,1,6), (8,2,6), (8,1,7)

Searching for (6,1,5) is not very useful, because we don't even know a solution to (6,1,6) !

Since I started this project 11 years ago, a lot of results were discovered, and quite a large amount of CPU has been spent onto them.

New results keep coming, but there is no real breakthrough.

Now, about the current Euler 625 project.

The program will compute solutions such that max(a,b,c,d,e,f,g)=7^6=117649

This may seem not very impressive, but don't forget that we will find solutions where the left or right numbers will be below

2*(117649^6)=5303461691719306943558046763202=5.303461691719306943558046763202e+30 !

or numbers using 103 bits.

My old project is able to compute (6,2,5) but it's very slow, compared to the algorithm used into this project.

The new project will help us double-check our old solutions (to check if we didn't miss one), and will probably find new ones.

I'll keep you informed here when a new solution is discovered.

Note also that the program discovers a lot of solutions where all the numbers are divisible by a small value (like 2, 3 or 5). We call these solutions not primitive, since a solution is still valid after a multiplication by any integer value !

Finally, I'd like to thank several people:

- Yoyo, because he setup the whole project. I hope we'll find new solutions very soon.

- Robert Gerbicz, who designed the new algorithm. Without him, we could not run this search

- Dead J. Dona for suggesting to start this project

- Greg Childers, who maintained the server for the old project during all these years, and finding new approaches.

Ok, I think that's all.

Good luck on this search !

The project is running since 11 years, but I think the new yoyo project will outperform our project in a few days.

The goal is to compute solutions to the equation:

a^6 + b^6 = c^6 + d^6 + e^6 + f^6 + g^6

We use the notation (6,2,5) to express the fact that the equation is at 6th power, has 2 left terms and 5 right terms.

Reminder: ^ means power. Thus a^6 = a*a*a*a*a*a.

For example, 40^6 = 4096000000, as you can see, a^6 grows very quickly.

Why is this equation particularly interesting ?

In fact, it is not very useful.

We hope to find a solution where one of the terms a,b,c,d,e,f or g is zero, in other words, this means that we are searching (6,1,5) or (6,2,4).

Why are we searching for (6,1,5) or (6,2,4) ?

Some mathematicians conjectured that we could find solutions of equations (k,m,n), where k=m+n.

In other words, we can find a combination of 6 terms at the 6th power that lead to 0 when added or subtracted.

Currently, we found solutions for the following equations:

(4,1,3), (4,2,2)

(5,1,4) (5,2,3)

(6,3,3)

(8,3,5) (8,4,4)

For example, for (6,3,3), Subba-Rao found the following result in 1934:

23^6+15^6+10^6=22^6+19^6+3^6

in fact, there is an infinite number of solutions for (6,3,3)

To extend the above list of results, the easier solutions to reach are, in increasing difficulty order:

(6,2,4), (7,3,4), (7,2,5), (7,1,6), (8,2,6), (8,1,7)

Searching for (6,1,5) is not very useful, because we don't even know a solution to (6,1,6) !

Since I started this project 11 years ago, a lot of results were discovered, and quite a large amount of CPU has been spent onto them.

New results keep coming, but there is no real breakthrough.

Now, about the current Euler 625 project.

The program will compute solutions such that max(a,b,c,d,e,f,g)=7^6=117649

This may seem not very impressive, but don't forget that we will find solutions where the left or right numbers will be below

2*(117649^6)=5303461691719306943558046763202=5.303461691719306943558046763202e+30 !

or numbers using 103 bits.

My old project is able to compute (6,2,5) but it's very slow, compared to the algorithm used into this project.

The new project will help us double-check our old solutions (to check if we didn't miss one), and will probably find new ones.

I'll keep you informed here when a new solution is discovered.

Note also that the program discovers a lot of solutions where all the numbers are divisible by a small value (like 2, 3 or 5). We call these solutions not primitive, since a solution is still valid after a multiplication by any integer value !

Finally, I'd like to thank several people:

- Yoyo, because he setup the whole project. I hope we'll find new solutions very soon.

- Robert Gerbicz, who designed the new algorithm. Without him, we could not run this search

- Dead J. Dona for suggesting to start this project

- Greg Childers, who maintained the server for the old project during all these years, and finding new approaches.

Ok, I think that's all.

Good luck on this search !