Sums of roots to approximate numbers I was thinking about random sequences, and thought up of this problem. 
Can algebraic or rational be expressed as a sum or difference of integer square roots? 
For example, $5=\sqrt{25}=\sqrt{16}+\sqrt{1}$.
It is known that if the series is infinite any number can be approximated, but is it possible to get any algebraic number(rational or irrational) with a FINITE number of terms? 
 A: Well certainly you cannot obtain a transcendental number as a sum of algebraics since algebraics are closed under sums. Nor can you obtain every algebraic number as a sum of radical integers (else every polynomial would be solvable!). But it's easy to see that one can obtain any rational integer.
Note: In fact Heine originally defined algebraic integers to be radical integers, i.e. the ring obtained by closing $\rm\:\mathbb Z\:$ under taking $\rm\:n$'th roots. Heine claimed that every solvable algebraic integer is a radical integer - a problem which is still open according to Franz Lemmermeyer. However, it is not too difficult to show that every quadratic integer is a radical integer, e.g.
$$ \frac{\sqrt{17}+1}{2}\ =\ \frac{\sqrt{17}+\sqrt{5}}{2}\ -\ \frac{\sqrt{5}-1}{2}\ =\:\ (7\ \sqrt{5} + 4\ \sqrt{17})^{1/3} - (\sqrt{5}-2)^{1/3}$$
A: Any finite sum or difference of integer square roots is an algebraic number, and $\pi$ is known to be transcendental.  So it is not possible
A: Nice question, that connects to a rich history (duplication of the cube, straight edge and compass construction of regular polygons).  It turns out, for example, that a root of a cubic with rational coefficients, which is irreducible over the rationals, cannot be expressed in the form you mention, or even more general form involving also division and nested square roots, like $\sqrt{\sqrt{5}+\sqrt{6}+17}$.  One can find proofs by searching on keywords like angle trisection, duplication of the cube.
