Tuesday, May 02, 2006

Someone Who's Good at Maths Help Me Out

OK, I'm not mathemetician. In fact, I've just found that I can't even spell the word.

But the other night I had a weird experience. I was thinking about something completely different, just dropping off to sleep, when I suddenly sat bolt upright, only vaguely aware of what I was doing, and blurted out: "3 cubed plus 4 cubed plus 5 cubed equals 6 cubed." Then I found I was incredibly alert and buzzing - couldn't possibly go back to sleep until I'd got out of bed and checked whether that calculation was accurate.

If I did my sums right, it is. But now I have loads of questions.

Is this something that is commonly known in maths circles? (Ha! Maths circles! So many possible jokes. None quite good enough to type here.)

Is there a general rule about come cubes being the some of three other cubes, the same way Pythagoras' Theorem applies to square numbers?

Have I just made an amazing discovery?

Can we call it "Joe's Theorem"?

Find out next week on The Saga of Mathematical Adventure, with idiot savant, Joe Craig.

Me.

2 comments:

Simon said...

Hi Joe,

It's a pretty observation, but don't give up writing books to go into number theory since
I'm afraid it has been noticed before (see mathworld.wolfram.com/
DiophantineEquation3rdPowers.html).

It fits into the study of 'Diophantine equations', which are equations containing only integers (whole numbers) where you look for integer solutions.

In this case it is an integer solution of
A^3 + B^3 + C^3 = D^3. There are many other solutions on the link above, but apparently there is no 'general solution', i.e. it is not known how to find all the solutions.

In general Diophantine maths can get extremely hard. For example, Fermat's Last Theorem says that there are no integer solutions to

A^n + B^n = C^n

for n>2. This was only proved recently (around 1995) using very advanced methods, several hundred years after it was first conjectured.

As for the Pythagoras' Theorem, if you go to three dimensions you are dealing with tetrahedra instead of triangles, but you still end up squaring things not cubing them. You get a^2 + b^2 + c^2 = d^2 where a,...,d are areas of sides of a special kind of tetrahedron (see www.cut-the-knot.org/Generalization/
pythagoras.shtml). So cubic formulae are not relevant.

Joe said...

Thanks, Simon.
Great response.
I too have always thought that "cubic forumlae are not relevant". It could be my mantra.
And doesn't 'Diophantine' sound like a type of precious stone? "Look, darling, I bought you a necklace. It's 24 carat Diophantine."