HMMT-Feb 2024 alg, Problem 8

Let $\zeta=\cos \frac{2 \pi}{13}+i \sin \frac{2 \pi}{13}$. Suppose $a>b>c>d$ are positive integers satisfying \[ \left|\zeta^{a}+\zeta^{b}+\zeta^{c}+\zeta^{d}\right|=\sqrt{3} \] Compute the smallest possible value of $1000 a+100 b+10 c+d$.

Solution: Solution: We may as well take $d=1$ and shift the other variables down by $d$ to get $\left|\zeta^{a^{\prime}}+\zeta^{b^{\prime}}+\zeta^{c^{\prime}}+1\right|=$ $\sqrt{3}$. Multiplying by its conjugate gives \[ \left(\zeta^{a^{\prime}}+\zeta^{b^{\prime}}+\zeta^{c^{\prime}}+1\right)\left(\zeta^{-a^{\prime}}+\zeta^{-b^{\prime}}+\zeta^{-c^{\prime}}+1\right)=3 \] Expanding, we get \[ 1+\sum_{x, y \in S, x \neq y} \zeta^{x-y}=0 \] where $S=\left\{a^{\prime}, b^{\prime}, c^{\prime}, 0\right\}$. This is the sum of 13 terms, which hints that $S-S$ should form a complete residue class mod 13 . We can prove this with the fact that the minimal polynomial of $\zeta$ is $1+x+x^{2}+\cdots+x^{12}$. The minimum possible value of $a^{\prime}$ is 6 , as otherwise every difference would be between -5 and $5 \bmod$ 13. Take $a^{\prime}=6$. If $b^{\prime} \leq 2$ then we couldn't form a difference of 3 in $S$, so $b^{\prime} \geq 3$. Moreover, $6-3=3-0$, so $3 \notin S$, so $b^{\prime}=4$ is the best possible. Then $c^{\prime}=1$ works. If $a^{\prime}=6, b^{\prime}=4$, and $c^{\prime}=1$, then $a=7, b=5, c=2$, and $d=1$, so the answer is 7521 . $\fbox{7521}$.

Answer: 7521