Japanese | Can you solve this? | Math Olympiad 

Interesting 🎉

Another great one. Thank you.

1,5

Congratulations

A = 5

a=3+2[cos(jπ/3)+i*sin(jπ/3)], j=0,1,...,5 and i=√(-1).

Solution 1: (a^2-6a+9)^3- 4^3=0 a^2-6a+9-4=0 (a-5)*(a-1)=0 a =1 or 5 Solution 2: (a^2-6a+9)^2+ 4(a^2-6a+9)+16=0 Set y =a^2-6a y^2+18y+81+4y+36+16=0 y^2+22y+133=0 (y+11)^2=-12 無理數 ！

5 куда уж проще.