# Need help proving an expression of roots of sums including roots

1. Jan 2, 2008

### santa

$$\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$$

2. Jan 2, 2008

### Hurkyl

Staff Emeritus
Done.

3. Jan 3, 2008

### robert Ihnot

One can try a lot of different things, like checking with a calculator to see if it really is close. (With Pari, a free program, you can check to 57 decimals.)

Anyway, you can let Y= (5+(21)^.5)^.5, and arrive at Y^4-10Y^2+4 = 0, and try to get the other side to do the same. But, really, there is a more straightforward approach.

What you really can do is show for both sides, Y^2 =5 + (21)^.5. The critical part is removing from under the radical:

$$\sqrt{8+4\sqrt3-2\sqrt7-\sqrt21}$$

Last edited: Jan 3, 2008
4. Jan 3, 2008

### dodo

I have a side observation. I was trying to multiply by conjugates to eliminate the roots. After an hour without getting nowhere, I was just treating the operands as vectors (a b c d), meaning $$a + b \sqrt 3 + c \sqrt 7 + d \sqrt 21$$, and noticing the operations behave as internal laws. There is some nice structure to it. With the bonus that cross products yield a different component of the vector, as it happens in complex math or with quaternions. Funny. It would be interesting to check what kind of structure is that. An algebraic integer field? It is not just a vector space, due to the cross products effect.

5. Jan 4, 2008

### robert Ihnot

Well, how about an algebratic extension of the ring of integers with a basis of:

$$1,\sqrt3, \sqrt7, \sqrt21$$?

6. Jan 5, 2008

### dodo

I'm happy someone put sense to my mumble-jumble. Also, I think that, if the coefficients were allowed to be rational, you could have multiplicative inverses, but so far I've been unable to derive an expression for it.

Specifically, I got stuck here,
$${a - b \bold i - c \bold j - d \bold k} \over {(a^2-3b^2-7c^2-21d^2) - (14 c d) \bold i - (6 b d) \bold j - (2 b c) \bold k}$$​
where i, j, k are resp. $$\sqrt 3$$, $$\sqrt 7$$, $$\sqrt 21$$.

Not that this will help solving the original problem, which has the sum of two roots as a complication. But it might ease the intermediate computations (of things inside roots) if you are doing them by hand.

Last edited: Jan 5, 2008
7. Jan 5, 2008

### Hurkyl

Staff Emeritus
Being a Q-vector space with an multiplication operation makes it a Q-algebra.

It's associative and has a multiplicative identity, so it's an extension ring of Q.

It is a field, which makes it an extension field of Q.

Each element is the root of a polynomial over Q, which makes it an algebraic extension of Q, and thus a number field.

It is finite dimensional over Q, which makes it an algebraic number field.

Incidentally, the easiest way to prove that it's a field is to find a way to embed it in C.

Last edited: Jan 5, 2008
8. Jan 5, 2008

### Hurkyl

Staff Emeritus
You know that your number field is a vector space over Q; you can leverage your knowledge of linear algebra to find inverses.

If you want the inverse of a number A, then "multiplication by A" is a linear transformation. You are thus trying to solve the linear equation
Ax = 1​

Another calculation is that, in this particular number field, each number has three conjugates. The product of all of the conjugates is a rational number:

$$\begin{equation*} \begin{split} &(a + b \sqrt{3} + c \sqrt{7} + d \sqrt{21}) \cdot (a - b \sqrt{3} + c \sqrt{7} - d \sqrt{21}) \\ & \cdot (a - b \sqrt{3} - c \sqrt{7} + d \sqrt{21}) \cdot (a + b \sqrt{3} - c \sqrt{7} - d \sqrt{21}) = q \end{split} \end{equation*}$$

Last edited: Jan 5, 2008
9. Jan 5, 2008

### dodo

That would mean inverting
$$\left( \begin{array} {cccc} a & 3b & 7c & 21d \\ b & a & 7d & 7c \\ c & 3d & a & 3b \\ d & c & b & a \end{array} \right)$$
for which... I could use Gauss elimination... or a math package. :D

The last paragraph about 3 conjugates might help me proceed with the denominator in post #6, where I got stuck earlier.

Ah, and thanks for post #7. This one I will keep for some time as a reference.

10. Jan 5, 2008

### Hurkyl

Staff Emeritus
Btw, I've made a change to reflect the usual terminology. (I'm pretty sure "algebraic extension of a ring" makes perfect sense, but the usual language refers to fields.

11. Jan 5, 2008

### santa

look this

12. Jan 5, 2008

### dodo

Clever! It is based on finding an integer n such that $$n (2 + \sqrt 3)$$, $$n (4 - \sqrt 7)$$ and $$n (5 + \sqrt 21)$$ are all squares of an element in the algebraic extension ring of post #5. I wonder if n=2 is the only possibility. Or if using different n 's with some simple gcd-like relation would also do.

It is a beautiluffy crafted problem, anyway.

P.S.: The first two squares must add to the third, too, in a sort of algebraic Pythagoras.
P.P.S: Oh sorry, I said another stupid thing (again). No Pythagoras here; it is the two numbers which add up, not their squares. That's why there is a sum of two roots in the original problem.

Last edited: Jan 5, 2008
13. Jan 5, 2008

### CompuChip

Well found, santa!

14. Jan 5, 2008

### robert Ihnot

CompuChip: Well found, santa!

$$\sqrt{8+4\sqrt3-2\sqrt7-\sqrt21} = \sqrt{(\sqrt3-1)^2/2*(\sqrt7-1)^2/2}$$ was how one could free the radical part after squaring left side.

Again, Wow! That procedure was really well-done santa! $$\sqrt {2 + \sqrt 3 } = \frac{{\sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }} = \frac{{\sqrt {\left( {1 + \sqrt 3 } \right)^2 } }}{{\sqrt 2 }} = \frac{{1 + \sqrt 3 }}{{\sqrt 2 }}.$$

Last edited: Jan 5, 2008
15. Jan 5, 2008

### dodo

I was in a number crunching mood, and found myriads of problems of the same sort, for the same base $$1, \sqrt 3, \sqrt 7, \sqrt 21$$, for example:
$$\sqrt{2-\sqrt 3}+\sqrt{11+\sqrt 21}=\sqrt{12+3\sqrt 7}$$​
$$\sqrt{4-\sqrt 7}+\sqrt{14+3\sqrt 3}=\sqrt{17+3\sqrt 21}$$​
$$\sqrt{107-15\sqrt 21}+\sqrt{14-5\sqrt 3}=\sqrt{96-9\sqrt 7}$$​

The original problem is, in a sense, "minimal": for the moment I can't find a similar equation where you don't allow coefficients in front of the $$\sqrt 3, \sqrt 7, \sqrt 21$$ (that is, if all non-real component of the vectors are limited to {-1,0,1}).

Last edited: Jan 6, 2008
16. Jan 5, 2008

### robert Ihnot

$$\sqrt{2-\sqrt 3}+\sqrt{11+\sqrt 21}=\sqrt{12+3\sqrt 7}$$

$$\sqrt\frac{(\sqrt3-1)^2}{2} +\sqrt\frac{(\sqrt21+1)^2}{2}= \sqrt\frac{(1+\sqrt7)^2}{2}*\sqrt3$$

Last edited: Jan 5, 2008
17. Jan 7, 2008

### disregardthat

It's relatively easy to find rationals a and b such that $$(a+b\sqrt{n})^2=c+d\sqrt{n}$$ for some given rationals c, d and n. (Assuming it exists) The root signs should then automatically be removed and the equality is obvious.

(Why do we have to type in [ tex ] notation [ /tex ] manually? Other forums have an own button for this such that when you highlight a part of your text, and then press the button it will automatically be sorrounded by the tex marks. I think that this forum should have the same option)

Last edited: Jan 7, 2008