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Solve this Maths question, or else, you aren't good enough for Oxbridge

Before you write bs, I know the answer

x+y+z= 1
x^2+y^2+z^2= 39
x^3+y^3+z^3= 61 In all respective integers

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13.


gg




(honestly, I could not care, so I stated my favourite number.)
Easy question, just takes a lot of time to do because of all the substitutions
Reply 3
Avatar for C=N-R
C=N-R
9
Original post by JAIYEKO
Before you write bs, I know the answer

x+y+z= 1
x^2+y^2+z^2= 39
x^3+y^3+z^3= 61 In all respective integers


So, you consider yourself to be some sort of elite mathematician?
Original post by JAIYEKO
Yes I do, problem?


I guess you could prove yourself by solving a problem my humble mind was capable of, then, rather than giving us a problem we cannot hope to solve.

So prove [0,1][0,1] and (0,1)(0,1) are homotopy equivalent, please.
Reply 5
Avatar for Zen-Ali
Zen-Ali
8
Original post by FireGarden
...


Bruh, ya dun dropped it.
BOVYRfl.gif1.2KB
Original post by JAIYEKO
Before you write bs, I know the answer

x+y+z= 1
x^2+y^2+z^2= 39
x^3+y^3+z^3= 61 In all respective integers


Damn, I got close, but my squares add up to 41 not 39 :frown:

[5, -4, 0]
Reply 7
Avatar for Aph
Aph
22
Original post by JAIYEKO
Before you write bs, I know the answer

x+y+z= 1
x^2+y^2+z^2= 39
x^3+y^3+z^3= 61 In all respective integers


-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

Posted from TSR Mobile
Original post by JAIYEKO
I'm arguably and possibly the only person in UK who can solve such of a difficult question


That's cute.
Reply 9
Avatar for Spairos
Spairos
13
Original post by JAIYEKO
Before you write bs, I know the answer

x+y+z= 1
x^2+y^2+z^2= 39
x^3+y^3+z^3= 61 In all respective integers


Not difficult at all.

The solutions are:
-0.5278597049, 4.910628066, -3.857842096
Original post by Aph
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

Posted from TSR Mobile


He said integers, so yo' wrong son.

Legendres three square theorem (correct me if I'm wrong) states that no three integers can add up to 39.
Reply 11
Avatar for slg60
slg60
17
Original post by JAIYEKO
I'm arguably and possibly the only person in UK who can solve such of a difficult question


See I told you that you are a troll.
There aren't integer solutions to this problem. Are you sure you didn't mean:

x^2+y^2+z^2=41?
Original post by TheFOMaster
He said integers


The solution I gave is the only one. That means that if he wants x,y and z to be integers then there is no solution.

OP, did you type the correct numbers?
Reply 14
Avatar for JAIYEKO
JAIYEKO
OP
Original post by Forum User
Damn, I got close, but my squares add up to 41 not 39 :frown:

[5, -4, 0]

The closest person yet
Reply 15
Avatar for JAIYEKO
JAIYEKO
OP
Original post by Aph
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

Posted from TSR Mobile

Integers, sorry
Reply 16
Avatar for JAIYEKO
JAIYEKO
OP
Original post by fat_hampster
There aren't integer solutions to this problem. Are you sure you didn't mean:

x^2+y^2+z^2=41?

I think i know what i wrote


For anyone that's interested ^
Screen Shot 2014-07-12 at 18.12.35.png124.5KB
Reply 18
Avatar for RichE
RichE
15
Original post by Aph
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

Posted from TSR Mobile


This - or one can show that it follows that xyz=1 so the roots can't be integers.
Original post by RichE
This - or one can show that it follows that xyz=1 so the roots can't be integers.


This one too.

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