Let $a,b$ be algebraic integers, i.e., the roots of polynomials with coefficients in $\mathbb{Z}$ and the leading coefficient is $1$.
Assume now that $a + b$ is equal to a rational integer--more specific $a + b = -1$.
What conditions must hold for $a$ and $b$? Is it possible to show that they both must be of degree $2$?
More specific: The two numbers are of the form $2 cos(\phi)$ for rational angles, and I want to count the number of solutions to the following equation depending on $n$ and $m$: $$ 2cos(\pi \frac{i}{n}) + 2cos(\pi \frac{j}{m}) = -1 $$ where $n,m \in \mathbb{N}$, $i \in \{1, \ldots,n-1\}$ and $j \in \{1, \ldots,m-1 \}$. So far I got this: Both summands have to be either rational or irrational. Nivens Theorem characterizes rational values of $cos$ at rational angels which handles the rational part.
So what about if both summands are irrational. We know that $2cos(\phi)$ are algebraic integers at rational angles. The golden ratio (times minus one) $-\frac{1+\sqrt{5}}{2}$ and its conjugate $-\frac{1-\sqrt{5}}{2}$ seem to be the only solutions here. They appear as values if $n = m$ and $n$ mod $5 = 4$ (two times).
One way to show this would be to bound the degree of $2cos(\pi \frac{i}{n})$. There are characterizations of these values for low degrees.
I am hoping that this was shown in some algebraic number theory textbook. My background in this field is not that great.
Let $a$ be any algebraic number (in particular, of any degree), and let $b=-1-a$. Then $b$ is also an algebraic number, and there is no reason to think it will be a conjugate of $a$. Also, no two algebraic numbers are algebraically independent (and no two numbers satisfying $a+b=-1$ are algebraically independent – that's an algebraic dependence relation, right there).
I think that answers most of your questions.
EDIT: The question has been edited to ask for solutions to $$2\cos(\pi j/m)+2\cos(\pi k/n)=-1$$ with positive integers $j,k,m,n$, with $j<m$ and $k<n$. The paper I mentioned in a comment, Conway and Jones, Trigonometric diophantine equations, Acta Arithmetica XXX (1976) 229-240, gives the answer. Using $2\cos x=e^{ix}+e^{-ix}$, the equation can be written as a vanishing sum of five roots of unity. Theorem 6 gives all the vanishing sums of nine or fewer roots of unity. The only sum of length five is $1+\beta+\beta^2+\beta^3+\beta^4$, where $\beta=e^{2\pi i/5}$. It's easy to get the answer from that.
But Theorem 7 is even better. It says,
Suppose we have at most four distinct rational multiples of $\pi$ lying strictly between $0$ and $\pi/2$ for which some rational linear combination of their cosines is rational but no proper subset has this property. Then the appropriate linear combination is proportional to one from the following list:
There follows a list of ten rational linear combinations of four or fewer cosines, but the only one of interest to us is the only one with exactly two cosines, and that is $\cos\pi/5-\cos2\pi/5=1/2$.
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The last section of Conway-Jones mentions "Crosby's equation", $$\cos2\pi a+\cos2\pi b+\cos2\pi c=0$$ which of course covers the equation in the current question---just take $c=1/3$. It gives the reference, W. J. R. Crosby, Solution to problem 4136, Amer. Math. Monthly 53 (1946) 103-107.