/sci/ - Science & Math

anus

x=0

27

x=0.198062

x= (1/3)

Let me rephrase the question ....
is there an integer value for 'x' which gives an integer value for the expression?

>>2913922

5.7161320270191653494103782635772243881529723332661869559892... -> FAIL

>>2913923

5.7161320270191653494103782635772243881529723332661869559892... -> FAIL

243

>>2913942

How does the square root of the square root of the square root of 1 equal 5.7?

>>2913948

hmm?

http://www.wolframalpha.com/input/?i=sqrt+%28sqrt%28sqrt%2827%29+%2B+27%29+%2B+27%29

2187

That's your answer, I guarantee it.

>>2913958

Um... 1/3 + 1/3 + 1/3 = 1, not 27

>>2913958
ITT 1==27

>>2913959

http://www.wolframalpha.com/input/?i=sqrt+%28sqrt%28sqrt%282187+%2B+2187+%2B+2187%29%29%29

Legit

Works with about 2.111857784835968374571940269
Wolfram Alpha, fags, do you use it?

>>2913971

see

>>2913939

>>2913969
Wolfram doesn't work with URL copypasta.
It got butchered.

>>2913959
>>2913959

Only correct answer in this thread

>>2913962

http://www.wolframalpha.com/input/?i=sqrt+%28sqrt%28sqrt%281%2F3%29+%2B+1%2F3%29+%2B+1%2F3%29

>>2913946
0,240,272,306,342,380,420,462,506,...

Now find a general rule for this sequence.

>>2913910
>243
no. 240 yes.

>>2913959
~47.2679884
nope

>>2913985

That's (sqrt 1/3) + (sqrt 1/3) + (sqrt 1/3), not sqrt(sqrt(sqrt(1/3 + 1/3 + 1/3)))

It doesn't take a fucking genius to figure out that 1/3 + 1/3 + 1/3 = 1

And then it doesn't matter how many square roots you put it through, sqrt 1 = sqrt sqrt 1 = sqrt sqrt sqrt 1 = 1

well let's start with 0.
Then you could solve the following equation:
(4n^4)x²-(4n^6+1)x+n^8=0
for each integer. It would give you some possible solutions.

>>2914014
wait no sorry

is there a formula or a rationale for reaching the solution or one must go by attempts?
math ignorant here.

the answer is zero, if you consider zero an integer, else there is no integer value for x for which the expresion sqrt(sqrt(sqrt(x)+x)+x) evaluates to an integer

>>2914062
>if you consider zero an integer

who doesn't consider zero an integer?!

penis

>>2913980
You are fucking retarded.
>HERP A DERP I DON'T LIKE DOSE OTHAR ANSWARS
They are still correct you humongous faggot.

>>2914078
this
>>2914062

and you're a moron. you need to look past 10.
>>2913997

>>2913997
nah dude, that's only for an infinite number of square roots

>>2914105
>herp derp derp derp
What? I do not speak derp.

The above referenced sequence is correct.

>>2913997
>Now find a general rule for this sequence.
There's an infinite number of functions which could start with this sequence.
I mean, literally infinite.
High school fags posting their shitty homework...

>>2914062
you could try the following: if n is an integer, you look for x such as
sqrt(x+sqrt(x+sqrt(x)))=n
which would imply (and only imply):
x+sqrt(x+sqrt(x))=n²
=>n²-x=sqrt(x+sqrt(x))
=>(n²-x)²-x=sqrt(x)
=>[(n²-x)²-x]²=x
=>x^4+x^3(-4n²-2)+ x²(6n^4+4n²+1)+x(-4n^6-2n^4-1)+n^8=0
then solve for every integer you want, and you would have to verify the solutions. That's the "reckless" method, I don't have any other idea for the moment.

so do we have any solution yet?

>>2913980
Is the "established" number 2187 supposed to be a value for x or the other integer

>>2914062
0 is in the set Z

I suspected OP was a fag but this adds further support.
The plot would reach the value 1/2 for any integers that met OP's condition.

There don't appear to be any from 1 to 10000.

I suspected OP was a fag but this adds further support.
The plot would reach the x axis for any integers that met OP's condition.

There don't appear to be any from 1 to 10000.

Maybe he just took the root from some nonexist-statement like fermats theorem

people seem confused about the function, this is how it looks

http://www.wolframalpha.com/input/?i=sqrt%28sqrt%28%28sqrt+x%29%2Bx%29%2Bx%29

>>2914150
I don't think there is a solution other than 0.

>>2914229
>>2914229
wtf? did you even understand the question? OP is not looking for integer values of x, he's looking for ANY value of x.

>>2913997
Um....none of these work... they're close, but they're not correct.

Well, 0 works.

>>2914241
wtf, did you even read >>2913939
OP wants only integers for x.

Again, 0 is probably the only solution.

>>2914241
>square root of a non integer that gives an integer
Let me know how that works out for you.

x~~0.198062, y~~1.

http://www.wolframalpha.com/input/?i=y%3Dsqrt%28sqrt%28%28sqrt+x%29%2Bx%29%2Bx%29%2C+y%3D1

If it was for any x, then take your pic. Any of the ones where the plot hits the x-axis would work.

>>2914269
Works well. See:
>>2914278

>>2914248
I'm sorry.
>>2914269
It's a bit more complicated than that, if you look closer.

38557890 is really close.

x = 90 almost works

>>2914292
Google is seemingly convinced.

>>2914310
6209.9999999967585128910939358617591132135625827455975516509

Pretty good find.

It can't be done.
Taking the inside part only.
RootX + X

Thats pretty much the same as X + X^2
I,e, X =5, then its
Root5 + 5
or X= Root5, then its
Root5 + (Root5)^2 which is the same.

For each square number you add an odd number to the previous square. 1 + 3 is 4, + 5 is 9 etc.

For this to work we need a number where X + X^2 has an integer as a square root.

If X = 6, then X^2 = 36, next square = + 13 to get 49. 13 > 6
If X = 12, then X^2 = 144, next square = + 25 to get 169, 25 > 12
If X = 230, then X^2 = 52900, next square = + 461 to get 53361, 461 > 230

No matter what value for X you put in you can never do X + X^2 and get a square number.
We've tried all values for X 1-10 already and know they don't work. So It's clear the only answer is X = 0.

Since there are no integer solutions to Root(RootX + X) there are obviously no integer solutions to the OP.

I'm 100% confident the only answer is 0.

There are no values for x which result in an integer besides 0, and here's why.

Denote OP's quantity as n. From the problem's premise, x is an integer, and n is also an integer, more specifically, non-negative integers. Begin solving for x.

sq(sq(sq(x) + x) + x) = n
sq(sq(x) + x) = n^2 - x
sq(x) + x = (n^2 - x)^2

Since x is a non-negative integer, denote x = k^2 with k a non-negative integer.

k + k^2 = (n^2 - k^2)^2

The right hand side of the equation is the square of an integer, and let's call that integer r.

k + k^2 = r^2.

The determinate of this equation in respect to k is sq(4r^2 + 1) = sq((2r)^2 + 1) which is never an integer unless r = 0, resulting to the trivial case of n = x = 0.

1

I would say there are arguments of irrationality that might help get the solution:
If x is an integer that matches the conditions,
sqrt(x+sqrt(x)) has to be an integer (esle, sqrt(x+sqrt(x+sqrt(x))) wouldn't be an integer)
then, x+sqrt(x) has to be an integer; therfore, x has to be a squared integer.

Say x=n² (n being an integer)
sqrt(n²+sqrt(n²+sqrt(n²)))=sqrt(n²+sqrt(n²+n)) is an integer.
which means that n²+n is a squared integer.
n²+n=y² so n=y²-n²=(y-n)(y+n) then n has to divide (y+n) (and automatically divides y-n because y-n=y+n-2n)
So y is a multiple of n. say y=q*n.
we have so far: n²+n=q²n².
Then: sqrt(n²+qn) is an integer.
so n²+qn is also a squared integer.
then we can write n²+qn=z²
=>qn=z²-n²=(z-n)(z+n)
n has to divide (z+n) (and z-n) so z= r*n
=>n²+qn=r²n²
and finally ((n+1)+n)n=sqrt(n²+sqrt(n²+sqrt(n²))).
which is (2n²+n)=sqrt(n²+sqrt(n²+sqrt(n²))).(I replaced q and r by their respective values because they depend on n)
then you just study the following functions:
x->2x²+x and x->sqrt(x²+sqrt(x²+sqrt(x²)))
and you see that the functions have an intersection in 0 and in a value in ]0;1[, and that's all.
so you conclude that the only integer that you are looking for is 0.

>>2914310
http://www.wolframalpha.com/input/?i=sqrt%28sqrt%28sqrt%2838557890%29%2B38557890%29%2B38557890%29

nope.jpg

Now take the derivative of that beast by hand.

http://www.wolframalpha.com/input/?i=sqrt%28sqrt%28%28sqrt%28x%29%2Bx%29%29%2Bx%29

For integer x, <span class="math">\sqrt{\sqrt{\sqrt{x}+x}+x}[/spoiler] is integer iff <span class="math">\sqrt{\sqrt{x}+x}[/spoiler] is integer iff <span class="math">\sqrt{x}[/spoiler] is integer. Hence <span class="math">x=p^2[/spoiler] for some integer p (this follows from the fundamental theorem of arithmetic).

So we know that <span class="math">\sqrt{p + p^2} = \sqrt{p(p+1)}[/spoiler] is integer. That is, <span class="math">p(p+1) = q^2[/spoiler] for some integer p and q. We can assume WLOG that p > 0, so q < p or q > p+1 are absurd. Since there are no integers between p and p+1, we must have either q = p or q = p+1.

The only solution is then q = 0, which leads to x = 0 or x = -1. By inspection, x = 0 is the only solution.

I have a feeling OP was fucking with us, given the file name.

>>2914409

there is no derivative because it's not defined as a function, x has a discrete value

>>2914428
2914398 here. I'm quite ashamed of not having thought of this. Good job man.
>>2914429: who cares, I find it to be quite an interesting problem to solve, so that's not a problem.

>>2914428
Oops, that shouldn't say iff but implies.

>>2914278

>mfw sine waves

Goddamn math, learn a new trick

>>2914390
I like this but what is this determinant you're speaking of?

>>2914476
The quadratic formula resolving in terms of k.

>>2914476
http://en.wikipedia.org/wiki/Determinant

>>2914480
Samefag, I meant discriminant, not determinant.

>>2914390
>>2914428
those are very close, and I think they're both good.

>>2914444
I was just plotting the imaginary portion of i^(2f[x]), which would be 0 at an integer. It just comes out in a sine pattern because, well, sine waves are everywhere.

x = GRAHAM'S NUMBER

OP here, this is a really elegant solution. I'm only in calc I and I came up with the question while playing with my calculator and finding roots of roots of i. The eighth root of i is <span class="math">\frac{\sqrt{\sqrt{\sqrt{2}+2}+2}}{2} + \frac{(\sqrt{\sqrt{\sqrt{2}+2}+2})i}{2}[/spoiler]

This led me to wonder about whether <span class="math">\sqrt{\sqrt{\sqrt{x}+x}+x}[/spoiler] had a solution that would ever result in an integer. Like an idiot I didn't think of 0, but after some fiddling, my intuition told me the problem had no solution. I knew if it did have a solution then x would have to be a perfect square, and so would root x plus x, ie that x^2+x would be a perfect square, but I wasn't smart enough to do much from there. I knew someone here would be able to figure it out though.

>>2914528

>>2914503

That's what I was saying! Math needs some new moves, mang.

>>2914428

OP here, this is a really elegant solution. I'm only in calc I and I came up with the question while playing with my calculator and finding roots of roots of i. The eighth root of i is <span class="math">\frac{\sqrt{\sqrt{\sqrt{2}+2}+2}}{2} + \frac{\sqrt{\sqrt{\sqrt{2}+2}+2}}{2}i[/spoiler]

This led me to wonder about whether <span class="math">\sqrt{\sqrt{\sqrt{x}+x}+x}[/spoiler] had a solution that would ever result in an integer. Like an idiot I didn't think of 0, but after some fiddling, my intuition told me the problem had no solution. I knew if it did have a solution then x would have to be a perfect square, and so would root x plus x, ie that x^2+x would be a perfect square, but I wasn't smart enough to do much from there. I knew someone here would be able to figure it out though.

>>2914428
I don't quite see why you can assume wlog that p>0?

Setting this equation equal to n and expanding leads to a long expression in form of;

2x^3 = ... n*n^3 + x*x^3 ...

Could we not employ now proven Fermat's Last Theorem ?

>>2914579
that's not really the good form, is it?

>>2914409

psh. integrate it by hand and then we'll talk.

I've been going through all the values of x for which the expression evaluates to an integer, going through every integer.
At some point it ends up approximating f(x)=x^2-x plus some value that very nearly approximates an integer reciprocal.
Conclusion: There is no integer solution for X such that f(x) is also an integer.

>>2914548

This function has no INTEGER solutions.

But the function is continuous and grows to infinity. <span class="math">\sqrt{\sqrt{\sqrt{x} + x} + x} = n[/spoiler] has a solution for every nonnegative integer <span class="math">n[/spoiler] by the Intermediate Value Theorem.