# 3rd Power Diophantine Generator

There is a generator to create perfect pythagorean triples…3:4:5, 5:12:13…
http://www.mathreference.com/num-zext,pt.html
for x^2+y^2=z^2,
x=u^2-v^2
y=2uv
z=u^2+v^2
when u and v are coprime.

For diophantine equasions of the third power
http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html
is there a similar generator?

Isn’t equation (8) on the linked page such a beast?

On the Wolfram Mathpage? Ugh, i got lost in the alphabet soup.:rolleyes:

I figured, to generate A^3+B^3+C^3=D^3 involved defining A,B,C, and D as functions of u,v,and w, as is the case with pythagorean triples.

Yeah equation (8) on the Wolfram page is:

(3 x^2 + 5 x y - 5 y^2)^3 + (4 x^2 - 4 x y + 6 y^2)^3 + (5 x^2 - 5 x y - 3 y^2)^3 = (6 x^2 - 4 x y + 4 y^2)^3

So let:
A = 3 x^2 + 5 x y - 5 y^2
B = 4 x^2 - 4 x y + 6 y^2
C = 5 x^2 - 5 x y - 3 y^2
D = 6 x^2 - 4 x y + 4 y^2

(with coprime x and y if you like)

Then A^3 + B^3 +C^3 = D^3.

The reason I was wondering is…well…I’d like to take an amateur crack at the Integer Brick problem:
http://www.durangobill.com/IntegerBrick.html
I figure, If I enter The 3rd power diophantine generator and the 2nd power diophantine (pythagorean triple) generator as variables in linear programming:
http://en.wikipedia.org/wiki/Linear_programming
http://vinci.inesc.pt/lp/
then this Great Unsolved Math Problem should become a simple, if tedious, algebra problem.

For that, you don’t want a 3rd power Diophantine equation; you want a 2nd power one with three terms. That is, you want integers a, b, and c such that a^2 + b^2 = d^2, b^2 + c^2 = e^2, a^2 + c^2 = f^2, and a^2 + b^2 + c^2 = g^2, where d, e, f, and g are all also integers. There’s no cubes here.

Sometimes you people just fuckin’ scare me.

OK.

According to this:
The generators for A^2+B^2+C^2=D^2 are

a=2mp
b=2np
c=p^2-(m^2+n^2)
d=p^2+(m^2+n^2)…given that m+n+p=(mod 2) and (m,n,p)=1
(whatever that means:confused::rolleyes:)

AND

a=2mp+2nq
b=2np-2mq
c=p^2+q^2-(m^2+n^2)
d=p^2+q^2+(m^2+n^2)

So…

If having the triple and quadruple generators, AND having

A^2+B^2=D^2
A^2+C^2=E^2
B^2+C^2=F^2
A^2+B^2+C^2=G^2

as the dimensions of the Brick, do we have enough information to input into a linear programming matrix?

You know you don’t even need the quadruple generator.

Combine:

A^2+B^2=D^2
and
A^2+B^2+C^2=G^2

To get:

D^2+C^2=G^2

Then you need to solve four pythagorean triples simultaneously:

A^2+B^2=D^2
A^2+C^2=E^2
B^2+C^2=F^2
D^2+C^2=G^2

In fact these are equivalent to equations (5) - (8) here -> (http://mathworld.wolfram.com/PerfectCuboid.html).

Can I ask you what it is that you are optimizing that suggests to you that linear programming will lead to a solution?

Also, which of the equations posted thus far do you feel are linear?

One way to “solve” the Integer Brick (for integral sides and an integral body diagonal, but not necessarily face diagonals) is to realize that the square of (a + b) is a^2 +2ab + b^2. So if two sides are a and b and you arrange that the square of the third side is 2ab = c^2, the body diagonal will be a + b.
The most obvious solution is a = 1, b = 2, c = 2, with diagonal 3 (and multiples of these), but this is not the only solution you can generate. You can also get a = 2 b = 9, c = 6 with diagonal 11, for instance. You can use this to generate an infinite number of solutions, not all of them multiples of lower-order solutions. The only trick is choosing the value of 2 ab = c^2, and the rest falls into place.

Since the problem could be stated like this:

I thought the solution could be arrived at with matrix mathematics, instead of linear programming:

but obviously you must first have digits to manipulate, not algebraic variables.

I halfway suspected that you were really trying to turn this into a linear algebra problem and that linear programming was a red herring. And as you’ve realized, linear algebra is used to solve (systems of) linear equations and is not really suited to this problem.

huh.

I guess the only thing left is a spreadsheet program with a TON of memory.

Brute-force exhaustive search of every pythagorean triple with digits from 1 to a quadrillion :eek:

I believe they already searched up to 1 trillion.

Not true – look at my post above. It is an infallible method of generating Pythagorean triples (quadruples?). And it doesn’t require that you test all possible triples.

I suspect that it won’t generate all possible integral pytrhagorean triples (quadruples?). But then again, the method you cite in your OP demonstrably doesn’t generate all Pythagorean triples of the form a^2 + b^2 = c^2.

Another option remains. You could prove that no solution exists. The advantage to this approach is that I strongly suspect that it is the only thing that has a non-zero probability of meeting success.

The OP’s method generates all Pythagorean triples in “lowest terms”; to generate all Pythagorean triples more generally, one just takes all the multiples of these.

AFAIK, that’s true. And it generates many of the multiples of the “lowest terms”, but not all of the multiples, as I’ve demonstrated myself. Although it generates all of the lowest Pythagorean triples i know of (at least in “lowest terms”), I must admit that I don’t know that it will generate all of the triples in “lowest terms”.

You don’t need a ton of memory, so long as you have some systematic way of going through the possible solutions. You don’t have to take up memory storing one that didn’t work; you just throw it out and move on to the next candidate. And each individual candidate takes up a small amount of memory.

Now, what you do need is a different way of managing your memory, since the standard integer data types aren’t large enough for numbers this big. But you can get packages to facilitate this; others have solved that problem before.

Gosh darn it. Last night, after a couple of drinks, I came up with a simple proof that no Integer Brick exists, QED.

It was so simple, I didn’t even bother to write it down. Needless to say, after a couple more, I wish I had. :smack: