<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] Calvin Lin provided an excellent proof. I am just adding in some details.
Obviously, $\sqrt{ 5} - \sqrt{2} > 0$
Also, $\sqrt{2} > 1$
$2 \sqrt{2} > 2$
$1 + 2 \sqrt{2} + 2 > 5$
$1^2 + 2 \sqrt{2} + (\sqrt{2})^2 > (\sqrt{5})^2$
$(1 + \sqrt{2})^2 > (\sqrt{5})^2$
Taking the positive square root, we have  $1 + \sqrt{2} > \sqrt{5}$
Then, $1 > \sqrt{5} - \sqrt{2}$
Combining the results together, we have $0 < \sqrt{ 5} - \sqrt{2} < 1$
Note that squaring a +ve quantity whose value lies between 0 and 1 is again positive but smaller.
Thus, $0 < (\sqrt{ 5} - \sqrt{2})^2 < \sqrt{ 5} - \sqrt{2} < 1$
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = … = 14$
$(\sqrt{5} + \sqrt{2})^2 = 14 - (\sqrt{5}- \sqrt{2})^2$
Therefore, $\lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13$
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