# Can six square numbers be simultaneously represented in a single sum of consecutive odd numbers? [closed]

I had some free time from my work to do a little exploration regarding the existence (or non existence) of perfect cuboids. A solution is represented by the set of Diophantine equations:

$a^2 + b^2 = x^2$

$a^2 + c^2 = y^2$

$b^2 + c^2 = z^2$

$a^2 + b^2 + c^2 = R^2$

If there exists a perfect cuboid, it implies that there is a consecutive sum of odd numbers from 1 to the $Rth$ odd number representing the space diagonal of a perfect cuboid, and that Six other perfect squares representing the face diagonals and edges also exist with the sum. It also means that each edge and face diagonal can be represented as a sum starting from 1 within $R^2$

I represent this consecutive arithmetic sequence like this:

$1+3+5+...+ R_o = R^2$ Where $R_o$ is the $Rth$ odd number in a consecutive sequence.

The set of edge, face diagonals and the body diagonals $[a,b,c,x,y,z,R]$ can each be represented similarly.

By using the fourth equation in the set, we see that:

$(1+3+5+...+a_o) + (1+3+5+...+b_o) + (1+3+5+...+c_o) = 1+3+5+...+R_o$

Then

$(1+3+5+...+x_o) + (1+3+5+...+c_o) = 1+3+5+...+R_o$

Since by first equation and our definition, $(1+3+5+...+a_o) + (1+3+5+...+b_o) = (1+3+5+...+x_o)$

Which Implies

$1+3+5+...+x_o = (c+1)_o + ... + R_o$

by subtracting $(1+3+5+...+c_o)$

Which implies:

$x^2 = (c+1)_o + ... + R_o$

By using the same process, we can derive this for the others:

$a^2 = (z+1)_o + ... + R_o$

$b^2 = (y+1)_o + ... + R_o$

$c^2 = (x+1)_o + ... + R_o$

$x^2 = (c+1)_o + ... + R_o$

$y^2 = (b+1)_o + ... + R_o$

$z^2 = (a+1)_o + ... + R_o$

Ultimately, the Perfect Cuboid Problem implies that there are two unique ways to represent each side and face diagonal in two different ways. Another way to see this is that it is possible to represent each side and face diagonal between $1+ ... + R_o$ both forwards and backwards. This also demonstrates that $a<b<c<x<y<z<R$

That is about as far as I got. My efforts now are attempts to demonstrate that it is not possible for one of these to be true, and by extension disproving others. I also believe that by extending the set of Diophantine equations, Perfect "Hypercuboids" can be disproved as well.

Suggestions are appreciated.

-

## closed as unclear what you're asking by Yemon Choi, Andrey Rekalo, André Henriques, Kevin P. Costello, David WhiteSep 10 '13 at 18:36

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

What is the question? –  Johan Wästlund Sep 10 '13 at 7:15
I vote to close as this is not a question. The only "question" comes at the end in the form "suggestions are appreciated", but these kind of questions are explicitely discouraged in the FAQ. –  André Henriques Sep 10 '13 at 8:02
This seems to me a reasonable question and I don't see why it should be closed. Maybe the questioner may improve the form according to the comments, stating more clearly the question (which is as I understand about non-existence of integer solutions), and removing the last sentence if it not appreciated (though it seems ok to me). –  Pietro Majer Sep 10 '13 at 8:23
The perfect cuboid problem is unlikely to fall to high school algebra. –  Gerry Myerson Sep 10 '13 at 12:59
I explicitly asked a question from the outset and provided some common courtesy at the end. If that is against the rules of this forum, especially when it comes to the discussion of open problem, I'll take it elsewhere. Thanks. –  user39814 Sep 10 '13 at 17:19