# How to prove an identity in radicals?

(4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))

$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$

How to prove it is right equality?

I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.

Any ideas?

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@hey, why don't you put as many close brackets as open brackets? I added them so that the question makes sense. –  anon Oct 3 '10 at 15:21
Please write something descriptive in the title of a question. –  Tsuyoshi Ito Oct 3 '10 at 15:24
@muad: nice edit, thanks. –  hey Oct 3 '10 at 15:31
LaTeX conversion of (16 / (14 - 6*sqrt(5))) - 6 = 2 * sqrt(61 + 24*sqrt(5)) : $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$ –  Américo Tavares Oct 3 '10 at 17:33
@hey, Suggestion: change title to "How to prove an algebraic numerical equality with radicals?" –  Américo Tavares Oct 3 '10 at 17:41

Simplify both sides to $8+6\sqrt{5}$, as they are both equal to this.

Just square the RHS to see this and rationalise the denominators on the LHS.

After the rationalisation of the denominators on the LHS (which is very quick) you obtain $(3+\sqrt{5})^2-6,$ and hence the $8+6\sqrt{5}.$

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Since

$$\dfrac{4^{2}}{\left( 3-\sqrt{5}\right) ^{2}}-\dfrac{\left( 6-5\sqrt{6}% \right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{4^{2}\left( 5-\sqrt{6}% \right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}}{% \left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}$$

and

$$\dfrac{1}{\left( 3-\sqrt{5}\right) ^{2}\left( 5-\sqrt{6}\right) ^{2}}=\dfrac{% \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}}{5776},$$

to prove

$$\left( \dfrac{4}{3-\sqrt{5}}\right) ^{2}-\dfrac{\left( 6-5\sqrt{6}\right) ^{2}}{\left( 5-\sqrt{6}\right) ^{2}}=2\sqrt{61+24\sqrt{5}}\quad (1)$$

it is enough to show that

$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}% \right) ^{2}$

$=11552\sqrt{61+24\sqrt{5}}$.

$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}% \right) ^{2}=3656\sqrt{5}+46208.$$

It remains to prove

$$\left( 34656\sqrt{5}+46208\right) ^{2}=11552^{2}\left( 61+24\sqrt{5}% \right) \quad (2).$$

The left hand side is

$$\left( 3656\sqrt{5}+46208\right) ^{2}=3202768896\sqrt{5}+8140370944$$

and the right hand side is equal:

$$11552^{2}\left( 61+24\sqrt{5}\right) =3202768896\sqrt{5}+8140370944,$$

which proves your equality.

$$\left( 16\left( 5-\sqrt{6}\right) ^{2}-\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}\right) \left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}$$

$$=16\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 5-\sqrt{% 6}\right) ^{2}-\left( 3+\sqrt{5}\right) ^{2}\left( 5+\sqrt{6}\right) ^{2}\left( 6-5\sqrt{6}\right) ^{2}\left( 3-\sqrt{5}\right) ^{2}$$

$$=5776\left( 3+\sqrt{5}\right) ^{2}-34656=34656\sqrt{5}+46208$$

Addendum 2: The second equality stated in the question

$$\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}\quad (3)$$

is equivalent to

$$\dfrac{8}{14-6\sqrt{5}}-3=\sqrt{61+24\sqrt{5}}=\sqrt{61+\sqrt{2880}}.$$

The LHS can be transformed into

$$\dfrac{8}{14-6\sqrt{5}}-3=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}$$

$$=\dfrac{18\sqrt{5}-34}{14-6\sqrt{5}}\times \dfrac{14+6\sqrt{5}}{14+6\sqrt{5}% }=\dfrac{64+48\sqrt{5}}{16}=4+3\sqrt{5}.$$

Then we have to prove

$$\sqrt{61+\sqrt{2880}}=4+3\sqrt{5}.\quad (4)$$

Now we apply to the LHS the following general transformation involving radicals:

$$\sqrt{A+\sqrt{B}}=\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}+\sqrt{\dfrac{A-\sqrt{% A^{2}-B}}{2}}$$

If $A=61,B=2880$, then $\sqrt{A^{2}-B}=\sqrt{61^{2}-2880}=29$ and

$$\sqrt{\dfrac{A+\sqrt{A^{2}-B}}{2}}=3\sqrt{5},$$

$$\sqrt{\dfrac{A-\sqrt{A^{2}-B}}{2}}=4,$$

which completes the proof.

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You're working too hard, Américo: simplify before you take all those squares! If we write the equation as a^2 - b^2 = 2Sqrt(c), begin by finding (as Derek Jennings suggests) that a = 3 + Sqrt(5) and b = -Sqrt(6). You should also suspect that c itself is a perfect square of the form x + y Sqrt(5); you easily find x = 4 and y = 3. –  whuber Oct 3 '10 at 16:17
whuber: I made an alternative "standard" algebraic computation (without any clever trick). –  Américo Tavares Oct 3 '10 at 17:25
@Américo Tavares: Just to be clear, my proposal is not a "clever trick;" it's a standard way to proceed. One is guided in such problems by seeking simplification early and often. This tends to limit the work that must be done. In fact, you could halve the length of your addendum by following this principle: simplify the arguments of the squares (and of the square root) before performing the squaring. –  whuber Oct 4 '10 at 14:17
@whuber: I agree with you that seeking simplification early and often is better than rarely and later. I also tend to do it, as opposed to what I did here. I am from a generation and a country where algebraic manipulation was very stressed, and we learned (as a problem) transformations of radicals such as the one I wrote in Addendum 2, no longer taught. I know that quite often all these manipulations are seen as boring and tedious, apart from being very time consuming. Also in trigonometry we studied many more formulae than now. In the future I will try more concise approaches. –  Américo Tavares Oct 4 '10 at 21:19
@whuber, only a final comment: The OP did not select my answer, which is a clear evidence to me that my answer was not good. –  Américo Tavares Oct 4 '10 at 21:29

You want to show:

$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$

First multiply out the squares

$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$

Then simplify it

$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$

now we can start getting rid of the square roots on the right hand side by squaring both sides

$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$

at this point it makes sense to rationalize the denominators of the fractions

$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$

which simplifies to

$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$

collecting like terms now gives

$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$

which is easily seem to be equal.

Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.

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You probably want to make sure both sides have the same sign before you square them. –  Oscar Cunningham Oct 3 '10 at 15:54
@muad: How do you center equations here? I don't know. –  Américo Tavares Oct 3 '10 at 21:29
Use $$eqn$$ instead of $eqn$. (Right-click + "Show Source" won't tell you that, but using the link menu at the bottom of the post, you can see it's numbered 5942, so go to math.stackexchange.com/posts/5942/revisions and you can view it.) –  Vandermonde Oct 3 '10 at 21:45
@freestorageaccount: Thanks! –  Américo Tavares Oct 4 '10 at 7:11

Specialize $\rm\ \ \ b=3,\ \ c = 5,\ \ d = 6 \ \Rightarrow\ a = 4\ \$ in this simple derivation:

$\rm\quad\quad\quad\quad\displaystyle \bigg(\frac{b^2-c}{b-\sqrt{c}}\bigg)^2 - \bigg(\frac{d -5\sqrt{d}}{5-\sqrt d}\bigg)^2$

$\rm\quad\quad =\quad \:(\: b \ \: + \: \sqrt{c}\ )^{\:2} \ \ \:-\ \ \ (\:-\:\sqrt{d}\:)^{2}$

$\rm\quad\quad =\ \ 2\ (\:a + b \sqrt{c}\ )\:, \quad 2\ a\ =\ b^2+c-d$

$\rm\quad\quad =\ \ 2\:\sqrt{a^2+b^2\:c+2\:a\:b\sqrt{c} }$

NOTE $\:$ Replacing numbers by functions makes the proof both simpler and more general - similar to your recently asked question

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Trivial. Lets simplify the left part: multiply first fraction to the sum of expressions in its denominator, and get whole part in second fraction and do with second fraction the same that we did with first:

$(\frac{4(3+\sqrt{5})}{9-5})^2 - (\frac{25-19 - \sqrt{6}}{5-\sqrt{6}}) = (3+\sqrt{5})^2 - (5-\frac{19(5+\sqrt{6})}{25-6})^2 = (3+\sqrt{5})^2 - 6 = 8 + 6\sqrt{5}$

Now lets prove this:

$8+6\sqrt{5}=2\sqrt{61+24\sqrt{5}}$

divide by two and square:

$4+3\sqrt{5} = \sqrt{61+24\sqrt{5}}$

$16+45+24\sqrt{5}=61+24\sqrt{5}$

Thats all.

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