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

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JAIYEKO
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#1
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#1
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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username1331498
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#2
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#2
13.


gg




(honestly, I could not care, so I stated my favourite number.)
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darkdrifter10
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#3
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Easy question, just takes a lot of time to do because of all the substitutions
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C=N-R
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(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?
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FireGarden
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(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] and (0,1) are homotopy equivalent, please.
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Zen-Ali
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(Original post by FireGarden)
...
Bruh, ya dun dropped it. Image
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Forum User
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(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

[5, -4, 0]
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Aph
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(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

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Manitude
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(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.
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Spairos
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(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
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TheFOMaster
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(Original post by Aph)
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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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.
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slg60
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(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.
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fat_hampster
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There aren't integer solutions to this problem. Are you sure you didn't mean:

x^2+y^2+z^2=41?
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Spairos
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#14
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(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?
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JAIYEKO
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#15
(Original post by Forum User)
Damn, I got close, but my squares add up to 41 not 39

[5, -4, 0]

The closest person yet
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JAIYEKO
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#16
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(Original post by Aph)
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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Integers, sorry
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JAIYEKO
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#17
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#17
(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
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Hunarench95
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#18
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Image

For anyone that's interested ^
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RichE
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#19
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#19
(Original post by Aph)
-3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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This - or one can show that it follows that xyz=1 so the roots can't be integers.
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Spairos
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#20
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(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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