+0

# See if you can solve this tough one! Show work.

+5
900
6

Suppose $$\sqrt{\sqrt{\sqrt[...]{\sqrt{\sqrt{x}}}}}=x^2-y$$,

for $$x,y\in \mathbb{Z},$$

for $$x \ge 0$$

Find the number of possible values for x+y.

Dec 16, 2015
edited by Guest  Dec 16, 2015

#1
+15

OMMGG, My tummy hurts just lookin' at that!

Dec 16, 2015

#1
+15

OMMGG, My tummy hurts just lookin' at that!

Hayley1 Dec 16, 2015
#2
+10

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

BOOOOOOOOOOOM!!!!! Dec 16, 2015
#3
0

ALAHUAKBAAARRR ???

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

Dec 16, 2015
#4
+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 so the 3 answers are   (0,0)   (1,0)  and  (1,2) So x+y  = 0 or 1 or 3

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

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

Your seconds question makes no sense to me. :( Dec 16, 2015
#6
0

Oops sorry Melody $$x=9y$$

Also can I have a proof?

Dec 16, 2015