# Strange Incredible Identity

1. Dec 9, 2007

### mathslover

How can we proceed to prove the following identity ?

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2. Dec 9, 2007

### Rogerio

The original expression :
$$\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}\ }\ }$$

First, note that 16=11+5 .

So, we can rewrite the last part :

$$\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{ ( \ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ ) ^2\ }$$

Or:

$$\sqrt{22+2\sqrt{5}\ }\ +\ \sqrt{5}\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ (\ \sqrt{11-2\sqrt{29}\ }\ +\ \sqrt{5}\ )$$

Thats the same as:

$$\sqrt{22+2\sqrt{5}\ }\ = \ \sqrt{11+2\sqrt{29}\ }\ +\ \sqrt{11-2\sqrt{29}\ }$$

Squaring both sides, we get:

$$22+2\sqrt{5}\ = 22\ +\ 2\sqrt{11^2\ -\ (2\sqrt{29})^2\ }$$

Or:

$$22+2\sqrt{5}\ = 22\ +\ 2\sqrt{121\ -\ 4*29\ }$$

Which is true.