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# Maths question watch

1. Hi everyone,

As 'Mathematical Moderator', here is your periodical maths puzzle.

It's not that difficult (my solution is about 3/4 of a page) and I'm interested in how people go about it:

Prove that it is impossible to factorise x^2 + y^2 + z^2 into linear factors. More precisely, prove that there do not exist real, or even complex, numbers a, b, c, d, e, f such that

Code:
`x^2 + y^2 + z^2 = (ax + by + cz)(dx + ey + fz)`
identically.

Now the interesting thing which arises out of this is that it seems we have a situation where both the real and even the complex fields are not enough to find constants to solve a simple problem involving only multiplication and addition. This draws parallels with the introduction of i = sqrt(-1) to form the complex field when it was discovered that certain polynomials could not be solved completely over the real field. Perhaps this question hints at the further extension of the complex field to include some other mystical number? What do you think? Can you define a 'number' with the required properties to find a contradiction to the above statement (i.e. find some new type of number which provides suitable a, b, c, d, e and f)? It's quite interesting, at least to me.

Regards,
2. x^2 + y^2 + z^2 = (ax + by + cz)(dx + ey + fz)
= adx^2 + bey^2 + cfz^2 + (ae + bd)xy + (af + cd)xz + (ce + bf)yz

Therefore:

ad = 1, be = 1, cf = 1
ae + bd = 0, af + cd = 0, ce + bf = 0

Using a = 1/d and ae + bd = 0 we have

e/d + bd = 0
e + bd^2 = 0

But b = 1/e

e + d^2/e = 0
e^2 + d^2 = 0

But this is only possible for d = e = 0, which means that there can be no x^2 or y^2 term. So it cannot work.
3. (Original post by mikesgt2)
x^2 + y^2 + z^2 = (ax + by + cz)(dx + ey + fz)
= adx^2 + bey^2 + cfz^2 + (ae + bd)xy + (af + cd)xz + (ce + bf)yz

Therefore:

ad = 1, be = 1, cf = 1
ae + bd = 0, af + cd = 0, ce + bf = 0

Using a = 1/d and ae + bd = 0 we have

e/d + bd = 0
e + bd^2 = 0

But b = 1/e

e + d^2/e = 0
e^2 + d^2 = 0

But this is only possible for d = e = 0, which means that there can be no x^2 or y^2 term. So it cannot work.
What about complex numbers?
4. (Original post by rahaydenuk)
Hi everyone,

As 'Mathematical Moderator', here is your periodical maths puzzle.

It's not that difficult (my solution is about 3/4 of a page) and I'm interested in how people go about it:

Prove that it is impossible to factorise x^2 + y^2 + z^2 into linear factors. More precisely, prove that there do not exist real, or even complex, numbers a, b, c, d, e, f such that

Code:
`x^2 + y^2 + z^2 = (ax + by + cz)(dx + ey + fz)`
identically.

Now the interesting thing which arises out of this is that it seems we have a situation where both the real and even the complex fields are not enough to find constants to solve a simple problem involving only multiplication and addition. This draws parallels with the introduction of i = sqrt(-1) to form the complex field when it was discovered that certain polynomials could not be solved completely over the real field. Perhaps this question hints at the further extension of the complex field to include some other mystical number? What do you think? Can you define a 'number' with the required properties to find a contradiction to the above statement (i.e. find some new type of number which provides suitable a, b, c, d, e and f)? It's quite interesting, at least to me.

Regards,
I can factorise, but not this...
5. (Original post by mikesgt2)
x^2 + y^2 + z^2 = (ax + by + cz)(dx + ey + fz)
= adx^2 + bey^2 + cfz^2 + (ae + bd)xy + (af + cd)xz + (ce + bf)yz

Therefore:

ad = 1, be = 1, cf = 1
ae + bd = 0, af + cd = 0, ce + bf = 0

Using a = 1/d and ae + bd = 0 we have

e/d + bd = 0
e + bd^2 = 0

But b = 1/e

e + d^2/e = 0
e^2 + d^2 = 0

But this is only possible for d = e = 0, which means that there can be no x^2 or y^2 term. So it cannot work.
As theone has mentioned, what about if we let e = i (= sqrt(-1)) and d = 1, this works!

Regards,
6. I don't have a solution, just a question to the solution suggested above.

Can't e^2+d^2=0 be possible
if e=i and d=1?

EDIT: didn't see the last entry
7. (Original post by haakon)
I don't have a solution, just a question to the solution suggested above.

Can't e^2+d^2=0 be possible
if e=i and d=1?
Yes, see above.
8. something went wrong when posting, ignore this
9. ae+bd=0

e/d+d/e=0

e^2 d^2 / de=0 then e^2 d^2=0 then de=0
then 2ed/de=0 is impossible

I'm not very good at maths, so i'm not sure if this is right.
10. (Original post by haakon)
ae+bd=0

e/d+d/e=0

e^2 d^2 / de=0 then e^2 d^2=0 then de=0
then 2ed/de=0 is impossible

I'm not very good at maths, so i'm not sure if this is right.
e/d + d/e = 0 is a correct implication, but then the implication (e^2 * d^2) / de = 0 is not I'm afraid. I think you probably meant (e^2 + d^2)/ed = 0, which doesn't help I'm afraid.
11. (Original post by rahaydenuk)
e/d + d/e = 0 is a correct implication, but then the implication (e^2 * d^2) / de = 0 is not I'm afraid. I think you probably meant (e^2 + d^2)/ed = 0, which doesn't help I'm afraid.
12. (Original post by integral_neo)
*Doesn't have a clue about any of this*
13. (Original post by integral_neo)
you have real numbers, with n = x
you can extend this to complex numbers, with n = x + iy
you then get quaternions, with n = x + iy + jz + ku
you can also get octonarions (or some name involving octo) with n = x + iy + ... (8 bases). these are useful in fundamental physics, read the NS article about it from last year.
14. yes i see what is happening here............

yes i also see what is happening here............

and here..................

but i don't understand a thing on the maths question..

i also see what is happening here......
15. (Original post by integral_neo)
16. I'll try it when I get home, but I don't know what a complex number is even so I doubt I'll get it!
17. This question can be done entirely without complex numbers I think
18. (Original post by theone)
This question can be done entirely without complex numbers I think
theone, haven't you solved this problem yet? ( i ddin't mean that in a patronizing way, i just thought you already solved it as you're so good at maths).
19. (Original post by bono)
theone, haven't you solved this problem yet? ( i ddin't mean that in a patronizing way, i just thought you already solved it as you're so good at maths).
Yeah, i solved it when it was initially set, i eventually found the solution, although i tried a few incorrect methods first.
20. So no one's got it yet? Hope Hayden comes out with it tomorrow, I'd love to see it.

Sod it, economics for me now. Damn I should've done that in the first place, might have got into Cambridge!

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