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Question Number 148559 by Jonathanwaweh last updated on 29/Jul/21

	Answered by Kamel last updated on 29/Jul/21
	$a=3k+r,b=3k′+r′ 0≤r<3, 0≤r′<3. a^2 +b^2 =3c=9(k^2 +k′^2 )+6(kr+k′r′)+r^2 +r′^2 ∴ r^2 +r′^2 =3(c−3(k^2 +k′^2 )−2(kr+k′r′)) So r^2 +r′^2 =3p, for (r,r′)∈{(0,1),(0,2),(1,2),(2,2),(1,1)} with symetry. r^2 +r′^2 ≢0[3] then r=r′=0 ⇒a≡0[3] and b≡0[3].$
	$a = 3 k + r, b = 3 k^{'} + r^{'} 0 ⩽ r < 3, 0 ⩽ r^{'} < 3 .$ $a^{2} + b^{2} = 3 c = 9 (k^{2} + k^{' 2}) + 6 (k r + k^{'} r^{'}) + r^{2} + r^{' 2}$ $∴ r^{2} + r^{' 2} = 3 (c - 3 (k^{2} + k^{' 2}) - 2 (k r + k^{'} r^{'}))$ $S o r^{2} + r^{' 2} = 3 p, f o r (r, r^{'}) \in {(0, 1), (0, 2), (1, 2), (2, 2), (1, 1)} w i t h s y m e t r y .$ $r^{2} + r^{' 2} ≢ 0 [3] t h e n r = r^{'} = 0 \Rightarrow a \equiv 0 [3] a n d b \equiv 0 [3] .$
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