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

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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
0
6 years ago
#2
13.

gg

(honestly, I could not care, so I stated my favourite number.)
1
6 years ago
#3
Easy question, just takes a lot of time to do because of all the substitutions
1
6 years ago
#4
(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?
1
6 years ago
#5
(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 and are homotopy equivalent, please.
0
6 years ago
#6
(Original post by FireGarden)
...
Bruh, ya dun dropped it.
0
6 years ago
#7
(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]
0
6 years ago
#8
(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
0
6 years ago
#9
(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.
6
6 years ago
#10
(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
0
6 years ago
#11
(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.
1
6 years ago
#12
(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.
0
6 years ago
#13
There aren't integer solutions to this problem. Are you sure you didn't mean:

x^2+y^2+z^2=41?
0
6 years ago
#14
(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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#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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#16
(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
0
#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
0
6 years ago
#18

For anyone that's interested ^
1
6 years ago
#19
(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.
0
6 years ago
#20
(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.
0
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