<s>[INST] Calculate the integer part I have to calculate the  integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
 I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
 [/INST] A direct way of solving this problem is to realize that 
$$ 0 < \sqrt{ 5} - \sqrt{2} < 1 $$
This can be seen by squaring both sides, and using that $ 6 < 2 \sqrt{10} < 7 $ which again is true by squaring.
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = 5 + 2\sqrt{10} + 2 + 5 - 2 \sqrt{10} + 2 = 14 $, hence this allows us to conclude that
$$ \lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13 $$
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