1. May 22, 2012

### autodidude

How would you approach something like this?

$$\sqrt{-\sqrt{3}+\sqrt{4+\sqrt{5}+\sqrt{17-4\sqrt{15}}}}$$

2. May 22, 2012

### e^(i Pi)+1=0

Try turning the radicals into exponents.

Last edited: May 22, 2012
3. May 22, 2012

### haruspex

Let $$17-4\sqrt{15} = x^2$$
There are two possibilities for the form of x:
$$x = A + B\sqrt{15}$$
$$x = A\sqrt{3} + B\sqrt{5}$$
The first gives 2AB = -4, A^2 + 15 B^2 = 17
I don't see any solutions to that. Try the other form ;-)

4. May 22, 2012

### Mandlebra

Set it equal to x, and repeatedly square. In the end you'll take the principal value of the root of whatever power x happens to have (think it might be 12).

5. May 22, 2012

### haruspex

If you go down that path you'll end up with a 32nd order polynomial. It almost certainly will not have integer roots, so I've no idea how you would go about finding factors.

(Update: as it happens it does have an integer root. The puzzle setter was unnecessarily kind.)

Last edited: May 23, 2012
6. May 22, 2012

### Mandlebra

Would you mind explaining why you choose the deepest radical?

7. May 22, 2012

### haruspex

I assumed that whoever set this puzzle knows there's a neat answer. The only chance I could see of finding a neat answer is to start with the innermost part and simplify that. Presumably that blends nicely with the next enclosing part, allowing the process to be repeated.
The answer, incidentally, is an integer.

8. May 24, 2012

### autodidude

@haruspex: I tried the second form and after a bit of algebra, I ended up with a negative square root (might've made a mistake somewhere or I might not've understood you correctly). Your method seems to be similar to what I was shown for evaluating expressions with radical within a radical (much simpler one though).

It was letting a+√b = √(4+2√(3)), squaring both sides and doing the algebra gave a much simpler expression of 1+√(3)

I used that method and tried experimenting with different variations of it (using a negative sign and adding different coefficients in front of √(b) - I'm not sure how they figured out that formula/method and am very interested in how these formulas are found) but my answer is always off by 2. My calculator says that the expression (the deepest one) should simplify to 2√(3) - √(5)

By the way, bit an elementary question, when we're squaring the equation, aren't we changing the original value?

9. May 24, 2012

### haruspex

I'm not aware of any mechanical process for solving these. Usually you can spot solutions. Remember that the numbers under the square roots in the answer are going to be factors of the numbers in the square roots of the original. And if there are two square roots in the answer, the numbers under them are going to be coprime. That's why, given √15, you know to try a + b√15 or a√3 + b√5:

3a^2 + 5b^2 = 17; 2ab = -4.
The second eqn establishes that a and b are some combination of +1, -2 or -1, +2.
The first settles it as a=2, b=-1 or a=-2, b=1.
a=-2, b=1 would give a negative value. By convention, real values of the √ function are taken to be positive.

Next step:
4 + √5 + 2√3 - √5 = 4 + 2√3 = (1+√3)^2

Final step:
-√3 + 1 + √3 = 1

10. May 25, 2012

### autodidude

Ah, thanks very much! Turns out it was a computational error on my part (ab = 4, rather than 2ab=-4...in my own defence, it was about 3 in the morning :p). I used the quadratic formula and got √(5/3) as well but quickly ended up with a negative root when I tried that. I guess you just have to be clever with these kinds of problems.

I found out they're called nested radicals but the stuff I found on how to solve them are way beyond me.