[sub-project] Perfect cuboid

[Zak]

O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.

[fwjmath]

O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.

Not really. It is possible that exactly two of them are odd, and by adding them up we have an integer again.

[Zak]

It is possible that exactly two of them are odd, and by adding them up we have an integer again.

In this case we can imaging that the sum of two odd values which we divide on 2 are equal any even value with dividing on 2. Thus we have in right part of equation sum of two even square values with common denominator 2, and sqrt of that will be irrational the same, I think.

[fwjmath]

It is possible that exactly two of them are odd, and by adding them up we have an integer again.

In this case we can imaging that the sum of two odd values which we divide on 2 are equal any even value with dividing on 2. Thus we have in right part of equation sum of two even values with common denominator 2, and sqrt of that will be irrational the same, I think.

Yes, but they may also be all divisible by 4, which makes their sum possible to be a square.

What you think is irrelevant, it is what you can prove that is important. I don't know what your math level is, but your reasonings so far clearly do not qualify even the university level, and we are dealing with a math research problem here that has troubled a lot of professional mathematicians.

[fwjmath]

It is possible that exactly two of them are odd, and by adding them up we have an integer again.

In this case we can imaging that the sum of two odd values which we divide on 2 are equal any even value with dividing on 2. Thus we have in right part of equation sum of two even square values with common denominator 2, and sqrt of that will be irrational the same, I think.

And for your convenience, please take a=b=1, c=4, g=3

and
a = 1, b = 1, c = 24
a = 1, b = 4, c = 1
a = 1, b = 4, c = 9
a = 1, b = 4, c = 15
a = 1, b = 4, c = 55
a = 1, b = 4, c = 89
a = 1, b = 7, c = 20
a = 1, b = 9, c = 4
a = 1, b = 9, c = 16
a = 1, b = 9, c = 40
a = 1, b = 15, c = 4
a = 1, b = 15, c = 32
a = 1, b = 15, c = 56
a = 1, b = 16, c = 9
a = 1, b = 16, c = 25
a = 1, b = 16, c = 79
a = 1, b = 20, c = 7
a = 1, b = 20, c = 39
a = 1, b = 20, c = 81
a = 1, b = 24, c = 1
a = 1, b = 24, c = 65
a = 1, b = 24, c = 71
a = 1, b = 25, c = 16
a = 1, b = 25, c = 36
a = 1, b = 32, c = 15
a = 1, b = 32, c = 55
a = 1, b = 36, c = 25
a = 1, b = 36, c = 49
a = 1, b = 39, c = 20
a = 1, b = 39, c = 64

Just to mention a few examples.

[Zak]

It is possible that exactly two of them are odd, and by adding them up we have an integer again.

can you give to me example of odd value of x:
x=y^2/2 ?
we must get an even value of sum of two such odd values, how you are approve
g = sqrt(d ^ 2/2 + e ^ 2/2 + f ^ 2/2), where d ^ 2/2 - odd and e ^ 2/2 - odd two, for example, how that possible?

[fwjmath]

can you give to me example of odd value of x:
x=y^2/2 ?
we must get an even value of sum of two such odd values, how you are approve

I don't need that. I have listed solutions to the equation 2*g^2 = a^2 + b^2 +c^2 above.

[Zak]

O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.

Not really. It is possible that exactly two of them are odd, and by adding them up we have an integer again.

You need. Give me example, what shows when at least d ^ 2/2 and e ^ 2/2 are odds

[fwjmath]

You need. Give me example, what shows when at least d ^ 2/2 and e ^ 2/2 are odds

As I said, d=e=1, f=4, g=3

[Zak]

d ^ 2/2 and e ^ 2/2 must be integger, because every odd value is integger: https://en.wikipedia.org/wiki/Parity_(mathematics)
this two expressions must be odd, what means integger too.

[fwjmath]

d ^ 2/2 and e ^ 2/2 must be integger, because every odd value is integger: https://en.wikipedia.org/wiki/Parity_(mathematics)
this two expressions must be odd, what means integger too.

They don't need to in order to get a perfect cuboid. You divide them by 2, because they appears twice in the sums, but that does not mean they divided by 2 must be an integer.

[Zak]

O.K.
So. We have an equation:
g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2
we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2.
Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give an irrational solution.

Not really. It is possible that exactly two of them are odd, and by adding them up we have an integer again.

You need give me an example to confirming your statement, otherwise mine is true and followed from Diophantine equations.