Perfect cuboid

The first three equations describe, according to the Pythagorean theorem, the lateral diagonals. And the fourth equation describes the main diagonal. I got the following result.

GCD(a, g)=1, GCD(b, g)=1, GCD(c, g)=1\\GCD(d, g)=1, GCD(e, g)=1, GCD(f, g)=1\\GCD(a, f)=1,GCD(b, e)=1,GCD(c, d)=1

Where GCD is the greatest common divisor.

Proof

Substituting the definition of 'f' into the fourth equation, we get

a^2+f^2=g^2

Similarly, we substitute the definitions of 'd' and 'e', ​​we get:

b^2+e^2=g^2\\c^2+d^2=g^2

We received the following system of equations:

\begin{cases}c^2+d^2=g^2(5)\\b^2+e^2=g^2(6)\\a^2+f^2=g^2(7 )\\a^2+b^2=d^2 (8)\end{cases}

It is easy to see that this system of equations is equivalent to the original one, just substitute 'd' from equation 8 into equation 5 and get equation 4.

\begin{cases}a^2+b^2+c^2=g^2\\b^2+e^2=g^2\\a^2+f^2=g^2\\a^ 2+b^2=d^2\end{cases}

Now let's substitute the definition of 'g' from equation 4 and equation 6. We get:

a^2+b^2+c^2=b^2+e^2\\a^2+c^2=e^2

We got the original equation 2. We can do the same thing and get equation 3. So, we have proven that the system of equations 5, 6, 7 and 8 is similar to the original one.

Now a few words and primitiveness. If there is a perfect cuboid, you can multiply its sides by a certain factor and get a larger cuboid. Or divide by a certain coefficient and get a smaller cuboid. So, a primitive cuboid is a cuboid whose sides cannot be divided by a natural number and a new smaller cuboid can be obtained. Which means that:

gcd(a,b,c)=1

This is not difficult to prove. If the gcd along the edges is equal to the number 't' (greater than 1), then the gcd of the diagonals is also equal to 't'. Because the diagonals are the sums of the squares of the edges. Now we can divide all sides by 't' and get a smaller cuboid, which contradicts the definition of a primitive cuboid.

So we have equation 5

c^2+d^2=g^2(5)

This is a Pythagorean triple! After all, a Pythagorean triple is when the sum of two numbers squared is equal to the third number squared. This is exactly the case here. And the Pythagorean triple is decomposed into a primitive triple multiplied by a certain coefficient. That is, equation 5 has the form

g^2=(K_1C_1)^2=(K_1A_1)^2+(K_1B_1)^2

Where:

GCD(A_1,B_1,C_1)=1

Once again about the primitiveness, but now the primitiveness of Pythagorean triplets. A Pythagorean triple is when the sum of the squared integers is equal to the third number squared. Similarly, if we multiply all three numbers by another natural number, we get a new Pythagorean triple. Well, or if we split it up. So, a primitive Pythagorean triple is when two numbers squared are equal to the third number squared. But they cannot be divided by some natural number and get a smaller Pythagorean triple.

Next it will be proven:

K_1=1

This means 'c', 'd' and 'g'. Form Primitive Pythagorean triple. This means that all three numbers are mutually prime to each other. This is where the result came from.

So we have:

g^2=(K_1C_1)^2=(K_1A_1)^2+(K_1B_1)^2(9)

Where

A_1,B_1,C_1-Primitive\component\\K_1-Coefficient

But we can also do equations 6 and 7. They also form a Pythagorean triple. Means:

b^2+e^2=g^2\\g^2=(K_2C_2)^2=(K_2A_2)^2+(K_2B_2)^2(10)\\a^2+f^2=g^2 \\g^2=(K_3C_3)^2=(K_3A_3)^2+(K_3B_3)^2(11)

If it is proven that all K are equal to 1. We get that in equations 5, 6 and 7 all numbers are pairwise coprime. This is where the result came from.

Now we note that since we are considering a primitive cuboid, then the gcd of all coefficients K is equal to 1. Because if it is not equal to 1, then all the side diagonals, edges and sides can be divided by it and get a smaller cuboid.

Means:

GCD(K_1,K_2,K_3)=1(12)

What we have:

g=K_1C_1=K_2C_2=K_3C_3

'g' is a certain integer, which we will represent three times as the hypothesis of a primitive Pythagorean triple, multiplied by a certain coefficient. This means that this number at least contains the LCM of these three hypotenuses (the LCM is the least common multiple). Means:

g=LCD(C_1,C_2,C_3)*K

And the LCM of any numbers is divisible by at least one of these numbers. Means:

g=t_1C_1K=t_2C_2K=t_3C_3K

It's easy to notice that:

K_i=K*t_i

But according to equation 12, the GCD of all K is equal to 1. This means:

GCD(K*t_1,K*t_2,K*t_3)=1

This means that K is equal to 1.

g=K_1C_1=K_2C_2=K_3C_3=NOK(C_1,C_2,C_3)

Means:

K_i=NOK(C_1,C_2,C_3)/C_i(13)

Now consider equation 8:

a^2+b^2=d^2

Let's substitute the definition of edges from equations 9, 10 and 11 into equation 8. We get:

(K_1A_1)^2 +(K_2A_2)^2=(K_3B_3)^2(14)

Let's substitute the value of the coefficient K_total from equation 13 into equation 14.

\frac{(LMK(C_1,C_2,C_3)*A_1)^2}{C_1^2}+\frac{(LMK(C_1,C_2,C_3)*A_2)^2}{C_2^2}=\frac {(NOK(C_1,C_2,C_3)*B_3)^2}{C_3^2}

We shorten everything to LOC by hypotenuse. We have:

\frac{A^2}{C_1^2}+\frac{A_2^2}{C_2^2}=\frac{B_3^2}{C_3^2}(15)

Let's multiply everything by C1:

A_1^2+\frac{C_1^2A_2^2}{C_2^2}=\frac{C_1^2B_3^2}{C_3^2}

On the right is a fraction, on the left is an integer + fraction. This means the denominators must be equal. Naturally after the reduction. Let me note that A2 and C2 are relatively prime (since they are the leg and hypotenuse of a primitive Pythagorean triple). This means they will not be cut back. So the denominators:

\frac{C_2}{GCD(C_1,C_2)}=\frac{C_3}{GCD(C_1,C_3)}

Multiply crosswise:

C_2GCD(C_1,C_3)=C_3GCD(C_1,C_2)(16)

Now we do the same step with the C2 coefficient. Let's multiply everything by C2:

\frac{C_2^2A_1^2}{C_1^2}+A_2^2=\frac{C_2^2B_3^2}{C_3^2}

I repeat: On the right is a fraction, on the left is an integer + fraction. This means the denominators must be equal. Naturally after the reduction. Let me note that A1 and C1 are relatively prime (since they are the leg and hypotenuse of a primitive Pythagorean triple). This means they will not be cut back. So the denominators:

\frac{C_1}{GCD(C_1,C_2)}=\frac{C_3}{GCD(C_2,C_3)}

Multiply crosswise:

C_1GCD(C_2,C_3)=C_3GCD(C_1,C_2)(17)

Equations 16 and 17 are equal, which means we can write:

C_1GCD(C_2,C_3)=C_2GCD(C_1,C_3)=C_3GCD(C_1,C_2)

Again we got a certain number that is divisible by any C-item. This means that this same number is divided by their LCM, the remaining number after division is written as K.

C_1GCD(C_2,C_3)=C_2GCD(C_1,C_3)=C_3GCD(C_1,C_2)=LCD(C_1,C_2,C_3)K(18)K_1C_1=K_2C_2=K_3C_3=NOK(C_1,C_2,C_3)

Means:

C_1GCD(C_2,C_3)=C_2GCD(C_1,C_3)=C_3GCD(C_1,C_2)=\\LCD(C_1,C_2,C_3)K=K_1C_1K=K_2C_2K=K_3C_3 K

Now we find out the value of the remaining coefficient K.

C_1GCD(C_2,C_3)=K_1C_1K

We reduce and express K1:

K_1=GCD(C_2,C_3)/K

Likewise:

K_2=GCD(C_1,C_3)/K\\K_3=GCD(C_1,C_2)/K

Just a little left 🙂

The GCD of all K-ities is equal to 1, which means:

GCD(GCD(C_2,C_3)/K,GCD(C_1,C_3)/K,GCD(C_1,C_2)/K)=1

Multiply both sides by K:

GCD(GCD(C_2,C_3),GCD(C_1,C_3),GCD(C_1,C_2))=K

Now let's remember the fundamental theorem of arithmetic. Any number can be represented as a product of prime numbers. This means that the GCD for three numbers takes the intersection of primes that are included in the decomposition into prime factors. Means:

K=GCD(C_1,C_2,C_3)

It turns out that equation 18 looks like:

C_1GCD(C_2,C_3)=C_2GCD(C_1,C_3)=C_3GCD(C_1,C_2)=\\GCD(C_1,C_2,C_3)GCD(C_1,C_2,C_3)(19)

Now consider an arbitrary prime number p. It is contained in all C-items, possibly to the degree 0. Let us denote the degrees in which they appear as α1, α2 and α3. Let's sort all C and all α.

So:

C_1\leq C_2\leq C_3\\α_1\leqα_2\leqα_3

Note that a prime to the power of alpha 1 is not necessarily included in C1. It may also be included in C3.

C_1GCD(C_2,C_3)=GLCD(C_1,C_2,C_3)GCD(C_1,C_2,C_3)(20)

Now a few words about LCM, GCD and their connection with the canonical expansion. The GCM takes the maximum degree for any prime, and the GCD takes the minimum.

Now we have 3 cases, our arbitrary prime p is contained in C1 either to the power of α1, or to the power of α2, or to the power of α3. Let's consider all three options.

1) Contained in the power α1, then for the prime power, equation 20 has the form:

α_1+α_2=α_1+α_3

On the left is alpha 2 because if α1 is included in C1, then only α2 and α3 remain, and GCD takes the minimum degree, that is, alpha 2.

2) Contained in the power α2, then for the prime power, equation 20 has the form:

α_2+α_1=α_1+α_3

Again, on the left is alpha 1, since if α2 is included in C1, then only α1 and α3 remain, and GCD takes the minimum degree, that is, alpha 1.

3) Contained in the power α3, then for the prime power, equation 20 has the form:

α_3+α_1=α_1+α_3

Now the equality is true!

This means that for any prime p, the degree to which it is included in C-ities, it is included in C1 to the maximum degree.

Means

C_3\leq C_1

It means that:

C_1=C_2=C_3

Means:

K_1=K_2=K_3

But the gcd of these numbers is equal to 1. And all three numbers are equal, which means that they are all equal to 1

K_1=K_2=K_3=1

Q.E.D.

This work relates to open problems of geometry. I am very grateful that you took the time to read this article. I hope that I was not mistaken.

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