NORVEGIAN
NORVEGIANĂThe 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.
