En identitet ikke så overraskende som det virker, fra magasinet FIBONACCI QUARTERLY / An identity not as surprising as it seems, from the magazine FIBONACCI QUARTERLY

NORVEGIAN

NORVEGIANĂ

          The identity

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

appears in the magazine FIBONACCI QUARTERLY, vol 12, no. 3 (1974) at pages 271 and 280.

          We will show a simple proof of this identity.

          It can be observed, with a little luck, that

$$16-2\sqrt{29}+2\sqrt{55-10\sqrt{29}}=$$

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

$$=\left (\sqrt{5}+\sqrt{11-2\sqrt{29}}\;\right )^2.$$

The right-hand side of the identity will then be equal to

$$\sqrt{11+2\sqrt{29}}+\sqrt{5}+\sqrt{11-2\sqrt{29}}\;.$$

Although it is usually said that "there is no formula for the sum of two radicals", we can boldly apply the formula $\sqrt{x}+\sqrt{y}=\sqrt{x+y+2\sqrt{x}\sqrt{y}}$, and obtain

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

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

$$=\sqrt{22+2\sqrt{11^2-4\cdot 29}}=\sqrt{22+2\sqrt{5}}.$$

     Then, the result of this calculation is $\sqrt{5}+\sqrt{22+2\sqrt{29}}$, which coincides with the left member of the identity

$\blacksquare$

          Remark 

          This solution was obtained by Mr. Constantin Telteu, to a post of mine on Facebook.