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)
$${a}=\mathrm{3}{k}+{r},{b}=\mathrm{3}{k}'+{r}'\:\mathrm{0}\leqslant{r}<\mathrm{3},\:\mathrm{0}\leqslant{r}'<\mathrm{3}. \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{3}{c}=\mathrm{9}\left({k}^{\mathrm{2}} +{k}'^{\mathrm{2}} \right)+\mathrm{6}\left({kr}+{k}'{r}'\right)+{r}^{\mathrm{2}} +{r}'^{\mathrm{2}} \\ $$$$\therefore\:{r}^{\mathrm{2}} +{r}'^{\mathrm{2}} =\mathrm{3}\left({c}−\mathrm{3}\left({k}^{\mathrm{2}} +{k}'^{\mathrm{2}} \right)−\mathrm{2}\left({kr}+{k}'{r}'\right)\right) \\ $$$${So}\:{r}^{\mathrm{2}} +{r}'^{\mathrm{2}} =\mathrm{3}{p},\:{for}\:\left({r},{r}'\right)\in\left\{\left(\mathrm{0},\mathrm{1}\right),\left(\mathrm{0},\mathrm{2}\right),\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{1},\mathrm{1}\right)\right\}\:{with}\:{symetry}. \\ $$$${r}^{\mathrm{2}} +{r}'^{\mathrm{2}} ≢\mathrm{0}\left[\mathrm{3}\right]\:{then}\:{r}={r}'=\mathrm{0}\:\Rightarrow{a}\equiv\mathrm{0}\left[\mathrm{3}\right]\:{and}\:{b}\equiv\mathrm{0}\left[\mathrm{3}\right]. \\ $$