JEE Advanced 2019 — Paper 1 — Question 41

Since $\omega$ is a cube root of unity, $\omega^3=1$, $1+\omega+\omega^2=0$, and $\overline{\omega}=\omega^2$. We compute $|a+b\omega+c\omega^2|^2 = (a+b\omega+c\omega^2)(a+b\overline{\omega}+c\overline{\omega}^2) = (a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$. Expand: $a^2+b^2+c^2+ab(\omega+\omega^2)+ac(\omega^2+\omega)+bc(\omega+\omega^2)=a^2+b^2+c^2-ab-bc-ca$. Simplify: $a^2+b^2+c^2-ab-bc-ca = \frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$. For distinct non-zero integers, the set of absolute differences $\{|a-b|,|b-c|,|c-a|\}$ is at least $\{1,1,2\}$ (e.g., for $a=1,b=2,c=3$). Thus the minimum value of $\frac12[(a-b)^2+(b-c)^2+(c-a)^2]$ is $\frac12(1^2+1^2+2^2)=\frac12(1+1+4)=3$. Hence the minimum is $3$.