Source: Kyiv mathematical festival 2021
Tags: Kyiv mathematical festival, number theory
Solve equation (3a−bc)(3b−ac)(3c−ab)=1000 in integers. (V.Brayman)
Taking equation (mod3) yields −a2b2c2 ≡1(mod3) and this we have a2b2c2 is ≡2(mod3) which is impossible. Thus there are no solutions.
In mod 3
(3a−bc)(3b−ac)(3c−ab)=1000≡(−bc)(−ac)(−ab)≡−a2b2c2≡1(mod3)
⟹a2b2c2≡2(mod3)
Note that if x2≡1;0(mod3)
⟹a2b2c2≡1;0(mod3)
which is impossible.