Question Number 148559 by Jonathanwaweh last updated on 29/Jul/21
Answered by Kamel last updated on 29/Jul/21
![a=3k+r,b=3k′+r′ 0≤r<3, 0≤r′<3. a^2 +b^2 =3c=9(k^2 +k′^2 )+6(kr+k′r′)+r^2 +r′^2 ∴ r^2 +r′^2 =3(c−3(k^2 +k′^2 )−2(kr+k′r′)) So r^2 +r′^2 =3p, for (r,r′)∈{(0,1),(0,2),(1,2),(2,2),(1,1)} with symetry. r^2 +r′^2 ≢0[3] then r=r′=0 ⇒a≡0[3] and b≡0[3].](Q148587.png)