Is $\sqrt 7$ the sum of roots of unity? Let $a_n$ and $b_n$ be 2 sequences of $n$ rationals. 
Is it possible that $\sqrt 7 = \sum_{m=1}^{n} a_m (-1)^{b_m}$ ? Is it possible that $\sqrt{17}$ = $\sum_{m=1}^{n} a_m (-1)^{b_m}$ ? 
How to prove or disprove these ?
 A: Let $\zeta=e^{2\pi i/7}$. We know that the sum
$$
1+\zeta+\zeta^2+\cdots+\zeta^6=0.
$$
Let
$$
S=\zeta+\zeta^2+\zeta^4.
$$
Then by squaring we get
$$
S^2=\zeta^2+\zeta^4+\zeta^8+2\zeta^3+2\zeta^5+2\zeta^6.
$$
Observe that $\zeta^8=\zeta$.
Subtract the above equation multiplied by two from this to get
$$
S^2=-2-\zeta-\zeta^2-\zeta^4=-2-S.
$$
Let $M=2S+1$. Then
$$
M^2=4S^2+4S+1=4(-2-S)+4S+1=-7.
$$
Therefore $M=\pm i\sqrt7$, and you can surely construct a sum of the required type from this.
The above recipe works for all primes $p$ instead of $p=7$ as long as you follow the rule that the exponents of $\zeta$ (here $1,2,4$) are the quadratic residues modulo $p$ (so you need $(p-1)/2$ terms in the sum $S$). Whether you get $+p$ or $-p$ as the square depends on the residue class of $p$ modulo $4$.
Look up Gauss' sums for details of the general case.
A: The square root of any number $n$, can be derived from the span of chords of a polygon $2n$, and $n$ if $4|n-1$.  Correspondingly, since the cords can be derived from cyclotomic roots of $1$ (ie numbers of the form $cis(\pi m/n)$ where $0<m<2n$, then all such square roots can be so derived. 
One approach to solving this sort of equation, is to look for a primitive root.  For $13$, we might note that $2$ is a primitive root.  We then create a power-series of $2$, modulo $13$.  This gives a series of 12 numbers.  We alternate these into two alternate series, eg 
$$c(n) = cis(2\pi n/13)$$
$$S1 = c(1)+c(4)+c(3)+c(12)+c(9)+c(10)$$
$$S2 = c(2)+c(8)+c(6)+c(11)+c(5)+c(7)$$
One can see that the subscript in $S2$ is twice that of $S1$, and the subscript of the following term in $S1$ is four times that of the base term.  So $3=4*4 mod 13$, etc.
Now, the isomorphism between $S1$ and $S2$ corresponds to a reversal of sign on a square-root.  There is a therom that the product of numbers $1-c(n)$ for $n$ = 1 to p-1, gives p (here $p=13$), so the square root here is that $13 = (6-S1)*(6-S2)$.  This works for all primes.
It should be noted that $6-S1$ is not exactly $\sqrt{13}$, but something that involves this prime, like $(13+3 \sqrt{13})/2$ or similar.  
A: Let $d$ be an integer. The field extension $\mathbb{Q}(\sqrt{d}) / \mathbb{Q}$ is a abelian extension, and therefore it is a subextension of a cyclotomic field extension $\mathbb{Q}(\zeta) / \mathbb{Q}$ where $\zeta$ is some root of unity. Thus, $\sqrt{d} \in \mathbb{Q}(\zeta)$ -- that is, $\sqrt{d}$ is a rational function of $\zeta$.
Since $\mathbb{Q}(\zeta)/\mathbb{Q}$ is an algebraic extension, this further means that $\sqrt{d}$ is a polynomial in $\zeta$ with rational coefficients, which is easy to put into the form you seek.
In fact, because $\sqrt{d}$ is an algebraic integer, it must lie in $\mathbb{Z}[\zeta]$, we can even select the $a_m$ to be integers. In fact, we can even arrange to have all of the $a_m$ be equal to 1, if we allow roots of unity to be repeated in the sum.
