Current Events > For which integers x is sqrt[x+sqrt(x+8)] a positive integer?

(+) Yeah! (+)

Note that the expression is a square root nestled in another square root.

You've got this fam

what is the practical real life application of the knowledge solving this problem confers

what socioeconomic advantage or benefit would come from knowing this

x = 1, 56, possibly something greater than 113

Do your own homework

1

Can't be fucked to figure out any others good luck.

yea that's a math problem got dang

is it literally just 1?

also 0

brb putting it in my graphing calculator

Robot2600 posted...

also 0

Zero doesn't work

Robot2600 posted...

is it literally just 1?

No.

Robot2600 posted...

also 0

The square root of 8 is not an integer, and the square root of that is even less so.

Infinite

Damn_Underscore posted...

Infinite

not an integer

Tyranthraxus posted...

not an integer

No there are infinite answers

Damn_Underscore posted...

No there are infinite answers

I'm gonna need you to prove that.

SQRT of SQRT of x + 8 = y

plug in any positive integer for y

Once you get past 255, the numbers wrap around anyway, so I guess you can call that an infinite amount of answers

Damn_Underscore posted...

SQRT of SQRT of x + 8 = y

plug in any positive integer for y

X also has to be an integer so you can't just "plug in any value"

squirt

Tyranthraxus posted...

X also has to be an integer so you can't just "plug in any value"

I guess thats true

Actually...

SQRT of SQRT of x + 8 = y

Square both sides

SQRT of x+ 8 = y

Square both sides

x + 8 = y

There B are infinite answer pairs for this

Damn_Underscore posted...

I guess thats true

Don't give up! If you plug in all real numbers into y and keep track of every value of x that is also an integer your method works!

Damn_Underscore posted...

Actually...

SQRT of SQRT of x + 8 = y

Square both sides

SQRT of x+ 8 = y

Square both sides

x + 8 = y

There B are infinite answer pairs for this

That's because the equation being asked about isn't the one that you just did.

im thinking just 1 then. i was trippin with 0 lol

Robot2600 posted...

im thinking just 1 then. i was trippin with 0 lol

Plug in 56 and see if it works.

ellis123 posted...

Don't give up! If you plug in all real numbers into y and keep track of every value of x that is also an integer your method works!

That's because the equation being asked about isn't the one that you just did.

Oh, I see now, I read it incorrectly

You can still do an algebra method though but Im not doing that

Decided to make a quick python script to brute force it. If I set this up right, I'm only getting 1 and 56 as answers. I've tried it up to 100 million values of X, and it still only gives 1 and 56.
https://gamefaqs.gamespot.com/a/user_image/3/3/9/AAVkBcAAD8R7.png
https://gamefaqs.gamespot.com/a/user_image/3/4/1/AAVkBcAAD8R9.png
That said, at 100 million values of x this script is starting to take awhile to run, so if there's some value beyond 100 mil that satisfies the condition then I won't be finding it lol

Damn_Underscore posted...

Actually...

SQRT of SQRT of x + 8 = y

Square both sides

SQRT of x+ 8 = y

Square both sides

x + 8 = y

There B are infinite answer pairs for this

The issue with this is once you get down to

y^2 = x + sqrt(x+8)

You're gonna have to square both sides, and you end up with
y^4 = [x + sqrt(x+8)]^2
And that leaves you with a mess:

y^4 = x^2 + 2*x*sqrt(x+8) + x + 8

Still have no idea what you would do with this to solve for only integers algebraically or if this is even the right direction

you can think about it and realize there is nothing else that will work because it's a very specific request. 64 is 2x2x2x2x2x2 so you can try other numbers like 512 and 4096 and realize it's probably not gonna work.

the top-level squareroot function means the answer is going to need a square of an integer, and that quickly goes to infinity that the +8 is too small to do anything, you'll always get an irrational squareroot because of that. squares aren't distributed like primes: 4, 9, 16, 25, 36, etc.... it always gets bigger and bigger to +8 will always fuck it up.

I read the problem wrong. If there was no "x +" the SQRT then that method would work fine

Robot2600 posted...

you can think about it and realize there is nothing else that will work because it's a very specific request. 64 is 2x2x2x2x2x2 so you can try other numbers like 512 and 4096 and realize it's probably not gonna work.

the top-level squareroot function means the answer is going to need a square of an integer, and that quickly goes to infinity that the +8 is too small to do anything, you'll always get an irrational squareroot because of that. squares aren't distributed like primes: 4, 9, 16, 25, 36, etc.... it always gets bigger and bigger to +8 will always fuck it up.

Yeah, you can see that intuitively, but is there any way to make a mathematical proof for it?

VampireCoyote posted...

squirt

This

Gwynevere posted...

Yeah, you can see that intuitively, but is there any way to make a mathematical proof for it?

Yes, but not easily. This is the type of problem that is very intuitive to humans that becomes very tricky to do a proof for.

Damn if I ly had I known that people on GameFAQs could do my calculus homework for me

Showing that the number do go to infinity is a proof, from what I've seen on numberphile.

Hmmm, if we replace the 8 with a 7, is the answer 42? Yea I think so. So that's why it's like that, you square the number and the. Subtract the number and that's the only thing that fits?

10 should be 90... Yup.

The 1 working was a fluke then? Seems so.

Rephrase the problem as:
Are there any perfect squares Y = X + 8 such that the expression Y + sqrt(Y) - 8 is also a perfect square?
Then just subtract 8 from Y to get back to X in the original problem.

So first and foremost, if sqrt(Y) = 8, have the obvious solution of Y + 8 - 8 = Y = a perfect square.

Perfect squares increase in how far apart they are by odd integers, which is twice as fast as the integers themselves increase, so you can very easily show that 64 is also the upper bound on this:
64 + 8 - 8 = 64, a perfect square, so Y = 64 and x = 56 works.
81 + 9 - 8 = 82, 1 more than itself and 18 less than the next perfect square
100 + 10 - 8 = 102, 2 more than itself and 19 less than the next perfect square
This pattern will continue, with all further integers getting farther from working on both sides.

I'm sure there's a smart algebraic way to solve this, but when you only have to test the squares of 1 through 8 for solutions after doing the above, you might as well just do that unless you're looking for more than just the set of answers. You'll find it works for 9 + 3 - 8 = 4 (Y = 9, X = 1) and nothing else before the Y = 64 we already had.

1 and 56 are indeed the solutions. In general, the number of positive integer solutions of sqrt[x+sqrt(x+c)] depends on the possible factorizations of 4c+1 .

So let c=11 and see how many solutions you get.

As someone who failed math in high school, what the fuck is an integer?

Vokrent posted...

As someone who failed math in high school, what the fuck is an integer?

Any real number with no decimal.

Tyranthraxus posted...

Any real number with no decimal.

There's numbers that aren't real?

Vokrent posted...

There's numbers that aren't real?

yea bruh, imaginary (make no mistake they are VERY real)

Vokrent posted...

There's numbers that aren't real?

Hyperreals, Surreal, and Complex numbers aren't Real or more accurately they're not part of a set that contains all real numbers.

Infinity is not a real number. 1.5 is real but not an integer.

I have more questions but I feel like I'm just gonna get a headache.

No further questions, your honour.

surreals are actually real numbers

complex(imaginary), irrational, and transcendental numbers are not

Robot2600 posted...

surreals are actually real numbers

complex(imaginary), irrational, and transcendental numbers are not

Irrationals are real.

Did I just start a math fight?

Tyranthraxus posted...

Irrationals are real.

so they are! look at that everyone was wrong but now we know.

Robot2600 posted...

so they are! look at that everyone was wrong but now we know.

Also surreal numbers aren't really a thing it's a surreal number set that includes reals, infinites, and infinitesimals.

you know what i meant, the ones cantor talked about.

I liked this one.

Answer:

We set sqrt(x+sqrt(x+8)) = a, for a some positive integer. Squaring out the roots then collecting coefficients, we get the equation:

0 = x^2 - (2a^2+1)x+(a^4-8)

Using the quadratic formula we notice that since (2a^2+1) is odd, (2a^2+1)^2-4(a^4-8) must be a perfect square and odd. The latter simplifies to 4a^2+33, so we set it equal to the square of an odd integer 2k+1 to get:

4a^2+33=(2k+1)^2

Subtract the (2k+1)^2 over and factor using difference of squares to get

(2a+2k+1)(2a-2k-1)+33=0

The product there must be an either an (11 and a 3) or a (33 and a 1), where one of them is negative. Solving this, the only positive solution to the former that works is when a=2 and a=8 for the latter. Going back to the polynomial in x^2, we solve and we get four solutions for x: 1, 8, 56, and 73. These are not guaranteed to solve the original equation, but all solutions to the equation are in that list, so you can check that the solutions are 1 and 56.