+0  
 
+5
565
6
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Suppose \(\sqrt[2016]{\sqrt[2015]{\sqrt[...]{\sqrt[3]{\sqrt{x}}}}}=x^2-y\),

for \(x,y\in \mathbb{Z},\)

for \(x \ge 0\)

Find the number of possible values for x+y.

Guest Dec 16, 2015
edited by Guest  Dec 16, 2015

Best Answer 

 #1
avatar+8613 
+15

OMMGG, My tummy hurts just lookin' at that!

Hayley1  Dec 16, 2015
 #1
avatar+8613 
+15
Best Answer

OMMGG, My tummy hurts just lookin' at that!

Hayley1  Dec 16, 2015
 #2
avatar+2592 
+10

Theoretically, it could be any positive number as X and Y are undefined and no negative can be under a square root.

BOOOOOOOOOOOM!!!!!

cheeky

SpawnofAngel  Dec 16, 2015
 #3
avatar
0

ALAHUAKBAAARRR ???

 

this one 1+x+1+1*3(1241:121343 = 92011224 find x!

Guest Dec 16, 2015
 #4
avatar+93691 
+5

Another way to write this is 

 

\(x=(x^2-y)^{2016!} \)

 

 

x and y must both be integers

 

If x=0 and y=0    then 0=0  that works

 

If   \(x=1\;\;and\;\;x^2-y=\pm1\)   

that would work to if they are both integers that is

1-y=1

y=0

so

if  x=1 and y=0  that works too

 

1-y=-1

2=y

so

if x=1 and y=2  that will work

and that is it     laugh

 

so the 3 answers are   (0,0)   (1,0)  and  (1,2)         cheeky

 

So x+y  = 0 or 1 or 3

 

Maybe another mathematician can do it a different / better way  ??

Melody  Dec 16, 2015
edited by Melody  Dec 16, 2015
edited by Melody  Dec 16, 2015
 #5
avatar+93691 
+5

 

Your seconds question makes no sense to me.    sad   :(angry

Melody  Dec 16, 2015
 #6
avatar+8 
0

Oops sorry Melody \(x=9y\)

Also can I have a proof?

Sonicmouse37  Dec 16, 2015

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