Robert Gerbicz hat geschrieben:Now reading that message and your first post I see what you don't understand.

What you want to do with Boinc is already known/searched... You want to search for g<2^32, where g is the space diagonal. But a^2+b^2+c^2=g^2, this means that (if X=odd term of {a,b,c}), X^2<g^2<2^64, so X<2^32. And this is known for more than 10(?) years. In fact all (primitive) Euler bricks is known for X<2^32,

http://arxiv.org/abs/math/0111229, what link I have already posted. Just to post one more data, my code would do it in about 1-2 days.

Robert Gerbicz,

I say again, you didn't understand my idea!

First 2 pages of this topic I spent to prove that space diagonal of primitive perfect cuboid is a product of primes 4k+1.

Do you exactly understand, what "a product of primes 4k+1" is?

A product of primes 4k+1 means that g = ? (p(i)^a(i)),

where ? means multiplication sign,

p(i) is prime of kind 4k+1,

a(i) is a power of p(i)

WE DON'T CHECK ANY OTHER ODD NUMBERS!!!

ONLY THOSE, WHO ARE A PRODUCT OF PRIMES 4k+1!!!

IF SOME ODD NUMBER HAS ANY 4k+3 FACTOR, IT WILL BE PASSED (IGNORED) AUTOMATICALLY!

BECAUSE WE WILL CHECK ONLY THOSE NUMBERS, WHO ARE PRODUCED BY MULTIPLICATION OF PRIMES 4K+1.

AND WE WILL NOT FACTORIZE ODD NUMBERS!!! WE

**WILL GENERATE** SPACE DIAGONALS BY MULTIPLICATION OF PRIMES! AND ONLY AFTER THAT WILL CHECK!

Every client will generate these numbers from the given set of primes 4k+1 in way described before!

The method of running over all these diagonals is described on the previous page.

Specially for you I'll list several first diagonals from the fisrt workunit:

g(1) = 5^1 = 5

g(2) = 5^2 = 25

g(3) = 5^3 = 125

g(4) = 5^4 = 625

g(5) = 5^5 = 3125

g(6) = 5^6 = 15625

g(7) = 5^7 = 78125

g(8) = 5^8 = 390625

g(9) = 5^9 = 1953125

g(10) = 5^10 = 9765625

g(11) = 5^11 = 48828125

g(12) = 5^12 = 244140625

g(13) = 5^13 = 1220703125

g(14) = 5^14 = 6103515625

g(15) = 5^15 = 30517578125

g(16) = 5^16 = 152587890625

g(17) = 5^17 = 762939453125

g(18) = 5^18 = 3814697265625

g(19) = 5^19 = 19073486328125

g(20) = 5^20 = 95367431640625

g(21) = 5^21 = 476837158203125

g(22) = 5^22 = 2384185791015625

g(23) = 5^23 = 11920928955078125

g(24) = 5^24 = 59604644775390625

g(25) = 5^25 = 298023223876953125

g(26) = 5^26 = 1490116119384765625

g(27) = 5^27 = 7450580596923828125

g(28) = 13^1*5^0 = 13

g(29) = 13^1*5^1 = 65

g(30) = 13^1*5^2 = 325

g(31) = 13^1*5^3 = 1625

g(32) = 13^1*5^4 = 8125

g(33) = 13^1*5^5 = 40625

g(34) = 13^1*5^6 = 203125

g(35) = 13^1*5^7 = 1015625

g(36) = 13^1*5^8 = 5078125

g(37) = 13^1*5^9 = 25390625

g(38) = 13^1*5^10 = 126953125

g(39) = 13^1*5^11 = 634765625

g(40) = 13^1*5^12 = 3173828125

g(41) = 13^1*5^13 = 15869140625

g(42) = 13^1*5^14 = 79345703125

g(43) = 13^1*5^15 = 396728515625

g(44) = 13^1*5^16 = 1983642578125

g(45) = 13^1*5^17 = 9918212890625

g(46) = 13^1*5^18 = 49591064453125

g(47) = 13^1*5^19 = 247955322265625

g(48) = 13^1*5^20 = 1239776611328125

g(49) = 13^1*5^21 = 6198883056640625

g(50) = 13^1*5^22 = 30994415283203125

g(51) = 13^1*5^23 = 154972076416015625

g(52) = 13^1*5^24 = 774860382080078125

g(53) = 13^1*5^25 = 3874301910400390625

g(54) = 13^2*5^0 = 169

g(55) = 13^2*5^1 = 845

and so on...

I hope it's clear how to generate every next diagonal from the previous, or not?

Any workunit will contain initial and finish states of $g$ described as a sets of primes 4k+1 with their powers.

Every workunit will have to check both big and small $g$

I've already attached the example of workunit.

I can repeat it here:

Code: Alles auswählen

```
[WU]
Ini=41^1*37^3*29^5*17^4
Fin=41^2*29^5*17^2*13^2*5^4
Amt=1995
[Primes]
5=1^2+2^2
13=2^2+3^2
17=1^2+3^2
29=2^2+5^2
37=1^2+6^2
41=4^2+5^2
```

Initial state is 41^1*37^3*29^5*17^4

finish state of workunit is 41^2*29^5*17^2*13^2*5^4

Every prime 4k+1, which will be necessary for workunit is described in Primes section with its decomposition to the sum of squares.

Do you understand why initial state 41^1*37^3*29^5*17^4 (3557744073931065217) formally is bigger than finish state 41^2*29^5*17^2*13^2*5^4 (1052500395367143125)?

Do you understand how to pass from 41^1*37^3*29^5*17^4 to 41^2*29^5*17^2*13^2*5^4, what is a sequence of diagonals you need to pass through, or explain?

If you yet don't, please read all my messages from the beginning again.

A client no needs any other information to execute this work unit.

Only initial state, finish state and decomposition of all primes, which will be necessary for him to pass through.

Robert Gerbicz hat geschrieben:
And from that message: "Will get 2'147'516'416 records."

And that would cost 16GB Ram (if you store it as two int values). That's too much, here.

My code would use less than 32MB Ram for X<1e13.

You have pulled out a phrase from a context.

I explained how to obtain the base of all primes with their unique decomposition for primes 4k+1.

All this work IS ALREADY DONE!

I ALREADY HAVE ALL PRIMES 2<p<2^32 WITH THEIR UNIQUE DECOMPOSITIONS!

SO A CLIENT WILL NOT NEED TO PRODUCE OR EVEN TO CHECK THESE NUMBERS TO PRIMARITY!

THEY WILL BE GIVEN TO A CLIENT AS IS, WITH THEIR DECOMPOSITIONS ALREADY DONE!

AND EVERY TIME ONLY A VERY SMALL SET OF THEM (maybe 6-7 at once). See example above.

A client will need only to compose different diagonals from the set of primes in way I described already several times,

and to check are there between diagonal decompositions the edges and face diagonals satisfying to conditions.

For your attention, my program would use less than 2MB Ram for ANY diagonal for ANY moment of runtime!

Robert Gerbicz hat geschrieben:ps. Reading again those messages, I'm not sure what you want to search. If in fact you want to cover g<2^64, then those tables would be much bigger. Or in other case I don't see the proof that g has no prime divisor larger than 2^32. This is a hole.

About g with divisor larger than 2^32 I'll share my thoughts later.

For today I don't see that you understand my idea basically.