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x3mEn hat geschrieben:Perfect Cuboid 1st Batch finish
Now we are close to 1st Batch ending. For those who's interested, what's next, we inform, after complete validation of 0-2^50 range we are going to publish a new app version which will be goaled to Perfect cuboids search only. Thus it's 7 times faster than the current is, so we expect that the next goal — 2^51 will be achieved quick enough.
Stay tuned!

Sounds nice.
If the new app is way faster you also also choose an wider testing range, whichs ends in less tasks needed to generate. That should also speed the search a bit up and will generate less work for the server.
(e.g. check enigma@home; due to very short tasks the server is very busy and often on high backlog or down due to high traffic)

Edit: Where do you expect to find a perfect cuboid? And where is your upper testing limit at all?

Aurel hat geschrieben: Edit: Where do you expect to find a perfect cuboid? And where is your upper testing limit at all?

To be honest, very unlikely a perfect cuboid exists at all, so I rather expect the project probable achievement will become the only short statement: "if a perfect cuboid exists, its space diagonal exceeds N". The only question, how high this N will, and how far we are able to rise. We'd already beaten the previous known to me result, 120 billion, almost 10'000 times. I believe we can raise it at least to 9P (~2^53).

For now the theoretical limit of my app is 2^63. With the future app version it needs appr. 1,25 mln years at a modern CPU. Or just 1 year continuos crunch of 150'000 modern 8-cores CPUs

x3mEn hat geschrieben: To be honest, very unlikely a perfect cuboid exists at all...

g^2=d^2+e^2+f^2-a^2-b^2-c^2=(a^2+b^2-c^2)+(a^2+c^2-b^2)+(b^2+c^2-a^2)=a^2+b^2+c^2
So you will snatch the jackpot when you will find:
If a<b<c then (a^2+b^2-c^2)<(a^2+c^2-b^2)<(b^2+c^2-a^2)
a^2=(a^2+b^2-c^2)
b^2=(a^2+c^2-b^2)
c^2=(b^2+c^2-a^2).
Is it possible?
PS.
The equation b^2=(a^2+c^2-b^2) has solutions when a<b<c, but neither a^2=(a^2+b^2-c^2) nor c^2=(b^2+c^2-a^2), all three equations have solutions only for a cube a=b=c but irrational diagonals
So we have 2 cases:
1. a^2+b^2 or/and a^2+c^2 or/and b^2+c^2 are not integer
2. g not integer

So

Hi! I found a proof:
The cube abc lies in a rectangular coordinate system so that one of its vertices is the origin, and its edges lie on the coordinate axes in positive directions, respectively. So: g^2=a^2+b^2+c^2 is the equation of a sphere of radius g with the center at the origin. Let a,b,c,g are integer. From the equation 2*g^2=d^2+e^2+f^2 we also have that there exists a larger sphere of radius g*√2 with the center at the origin, where face diagonals d,e,f are coordinates in a given coordinate system. Suppose that d,e,f are integer and can be expressed by rational fractions with the help of the coordinates a, b, c, thus: a/d, b/e, c/f must have rational values. But this is impossible, since the equation of a sphere of such a radius in a given coordinate system on the coordinate axes and centered at the origin will also be described by the formula: 2*g^2=(a*√2)^2+(b*√2)^2+(c*√2)^2, where: a*√2, b*√2, c*√2 are coordinates on the respective axes and √2 is irrational value. So our assumption is uncorrect and d,e,f together cannot have integer values!

Hi! How is your searching? )))

So I mean it's impossible to reduce left part with a sum of three radicals to right part which consist of single radical of the finaly expression:

x3mEn
Try to simplify the expression: sqr(a^2+b^2)+sqr(a^2+c^2)+sqr(b^2+c^2) to the desired form at the right part: q*sqr(a^{2}+b^{2}+c^{2})
Maybe you even get also to simplify sqr(u)+sqr(v)+sqr(w) which united to the single common radical
Try to express the variable n only through variables u,v,w, and later we can define the value of q.

x3mEn
I will help you next: Sorry, mistake was corrected.
Now everything is checked and true.

If I was not mistaken, then it's followed: