TITLE: Is $\sqrt 7$ the sum of roots of unity?
QUESTION [19 upvotes]: 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 ?

REPLY [2 votes]: 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.