Prove: there is a unique pair of integer roots of unity which differ in real part by $1$. I saw the following lemma somewhere, and I hope I did not misread it:
If $z_1$ and $z_2$ are $n$th and $m$th roots of unity respectively ($n,m$ positive integers possible equal), and the real part of $z_1-z_2$ is $1$, then the pair $(z_1,z_2)$ is unique.

This is really equivalent to showing the difference of some cosines is 1, which by the sum to product formula is equivalent to showing the product of two sines is $-1/2$, but I am not sure how to proceed from there.
 A: There are multiple solutions to the way the lemma as you remember it is phrased.
The primitive fourth root of unity, $i$, has real part differing from the primitive second root of unity, $-1$, by an amount of 1.  $\Re(i-(-1))=1$, so $(i,-1)$ is a solution.
Another solution which uses primitive roots, is the primitive sixth root compared to the primitive third root.  You have $\Re(e^{\frac{2\pi i}{6}}-e^{\frac{2\pi i}{3}})=\Re((\frac{1}{2} + \frac{i\sqrt{3}}{2})-(-\frac{1}{2}+\frac{i\sqrt{3}}{2})) = \Re(1) = 1$, so $(e^{2\pi i/6}, e^{2\pi i/3})$ is another solution different from the first.
If you were to restrict your attention only to primitive integer roots of unity, these (and using the primitive first root compared to the primitive fourth root) are likely to be the only examples, as a primitive $n^{th}$ root of unity with $n> 4$ will have positive real part, as will a primitive $m^{th}$ root of unity.  Positive number less than or equal to 1 minus a positive number less than or equal to 1 will necessarily be less than 1.
If you do not restrict yourself to using primitive roots only, then there are likely to be many more solutions (likely infinitely many, though I haven't yet thought of a proof).
