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

Question

How to prove the following identity?

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

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?

Answer 1

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 $8+6\sqrt{5}.$

Answer 2

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.

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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.

Answer 3

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

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

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

But (see addendum)

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

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.

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Addendum:

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

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

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

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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.

Answer 4

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.

Answer 5

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