# 3-D Pythagorean Theorem?

1. Jun 30, 2004

### dperez3894

Has anybody else tried this?

a^3 + b^3 + c^3 = d^3

3^3 + 4^3 + 5^3 = 6^3

27 + 64 + 125 = 216

This seems to be a logical extension of the Pythagorean Theorem and it works if the values of 3, 4 and 5 are used for a, b and c.

Has this already been discovered in mathematics or is this something new?

2. Jun 30, 2004

### jcsd

The logical extension of the Pythagorean theorum in 3 dimensions is

s^2 = x^2 + y^2 + z^2

3. Jun 30, 2004

### matt grime

The term 3d pythagoras is usually reserved to mean that the square of the length of a diagonal of a cube is the sum of the squares of the sides.

What is your theorem anyway? I only see an example that you've found some numbers whose cubes are related in a certain way.

4. Jun 30, 2004

### uart

Yeah that's kind of interesting, Fermat's famous conjecture was that there exist no equivalent of Pythagorean Triads for powers higher than two, eg no chance for integers a^3 + b^3 = c^3.

So what you're saying is that although there is no direct cubic "triad" equivalent there are indeed integer "cubic quartets". Interesting idea, perhaps there are also 4th power "quintets" and fifth power "sextet" etc. Does anyone know if there are existing theorems or conjectures about this?

5. Jun 30, 2004

### Muzza

Yes, Euler conjectured that there were no integers x, y, z, w such that x^4 + y^4 + z^4 = w^4 (not exactly what you were asking for, but close enough). Noam Elkies of Harvard discovered this counterexample in 1988:

2682440^4 + 15365639^4 + 18796760^4 = 20615673^4.

Last edited: Jun 30, 2004
6. Jun 30, 2004

### matt grime

And there is the famous example in this vein that every integer (and hence every square, cube 4th power etc) is the sum of 4 squares.

7. Jun 30, 2004

### robphy

Last edited by a moderator: Apr 21, 2017
8. Jun 30, 2004

### matt grime

"This seems to be a logical extension of the Pythagorean Theorem and it works if the values of 3, 4 and 5 are used for a, b and c."

but doesn't if a=b=c=1

9. Jun 30, 2004

### Digit

The more general question is: which sums are products?
A^3 = A^2 + A^2 + A^2 or A^3 = 3*A^2
From that A must equal 3, or for general:
N^n = n*N^n-1

So for any two sums like Z^n = X^n + Y^n
n can only be two.
Just started messing with this. Don't know where it goes.

10. Jun 30, 2004

### fourier jr

There's an n-dimensional Pythagorean theorem too isn't there? I don't see why not. How about a_1^2 + a_2^2 + .... + a_n^2 = a^2

11. Jun 30, 2004

### matt grime

Is that the shortest known "proof" of Fermat's last theorem?

12. Jun 30, 2004

### matt grime

Yes and no. The n dimensional version is a direct consequence of the 2d version; it is provable directly from it. Of course one might argue that this is just a formal result from making the definitions of inner products such as they are, though I must ask, is no one else actually going to say what any of the terms in their 'theorems' actually are? Pythagoras DOES NOT say that x**2+y**2=z**2, since 1,1,3 for x,y,z resp disproves that (even if we assume x,y,z must be real numbers in the first place!) it states something geometrical. Is the OP going to state what they might actually mean?

Last edited: Jun 30, 2004
13. Jun 30, 2004

### cair0

x^2 + y^2 = s^2

s^2 + z^2 = r^2

x^2 + y^2 + z^2 = r^2

14. Jul 1, 2004

### Digit

I don't know. I am sure there is a short proof but I don't know how to do it.

15. Jul 1, 2004

### matt grime

Your above post claims you do though.

16. Jul 1, 2004

### Gokul43201

Staff Emeritus
This (the first 3 lines are okay) doesn't make any sense to me. Can someone (Digit?) please explain ?

17. Jul 1, 2004

Staff Emeritus
Euler showed that the product of two sums of four squares is again a sum of four squares. This was part of his proof that every integer is the sum of four squares (including squares of zero where necessary). He did it by working out all the partial products and collecting terms.

18. Jul 2, 2004

### Ethereal

How was the counter-example discovered? By the use of computers?

19. Jul 2, 2004

### Muzza

Probably. See Noam Elkies' article "On A^4 + B^4 + C^4 = D^4, Math. of Comp. 51 (Oct. 1988), 825-835". ;)

20. Jul 3, 2004

### uart

My guess is it was a computer search, it's quite large seach to find those numbers, of the order of the largest LHS number to the forth if you do it by "brute force".

I wonder if anyone has tried searching for a fifth order example, a^5 + b^5 + c^5 + d^5 + e^5 = f^5 ?