a ∈ N. a is not a multiple of 3. 1) Show that a^3 ≡−1[9] or a^3 ≡1[9]. 2) Given a; b; c ∈ Z. Deduct from 1) that if a^3 +b^3 +c^3 ≡0[9] , then one of integers a; b; c is divisible by 3.


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Question Number 119575 by mathocean1 last updated on 25/Oct/20

$a \in N . a is not a multiple of 3 .$ $1) Show that a^{3} \equiv - 1 [9] or a^{3} \equiv 1 [9] .$ $2) Given a; b; c \in Z .$ $Deduct from 1) that if a^{3} + b^{3} + c^{3} \equiv 0 [9], then$ $one of integers a; b; c is divisible by 3 .$
	Answered by mindispower last updated on 25/Oct/20

	$a = 3 k \underline{+} 1$ $\Rightarrow a^{3} {(3 k \underline{+} 1)}^{3} = 27 k^{3} \underline{+} 27 k^{2} + 9 k \underline{+} 1$ $= 9 (3 k^{2} \underline{+} 3 k^{2} + k) \underline{+} 1$ $\Rightarrow a^{3} = \underline{+} 1 [9]$ $a^{3} + b^{3} + c^{3} \equiv 0 [9]$ $s u p p o s e t h a t n o n o f t h e m \equiv 0 [9]$ $\Rightarrow a^{3}, b^{3}, c^{3} \equiv \underline{+} 1 [9]$ $a^{3} + b^{3} + c^{3} \equiv (- 3, - 1, 1, 3) [9]$ $a r e a l l p o s s i b i l i t y \neq 0 [9]$ $\Rightarrow o n e o f t h e m e \equiv 0 [9]$ $e x e m p l e a = 3 k, b = 3 e + 1, c = 3 w - 1$
	Commented by mathocean1 last updated on 25/Oct/20

	$please sir at the eighth line can you detail$ $how - 3; - 1; 1; 3 appeared ?$
	Commented by mindispower last updated on 25/Oct/20
	$yeah withe pleasur since evrey a^3 =+_− (9) what is possible value of x+y+z withs z,y,x∈{−1,1} if one =−1⇒−1+1+1=1 if two=−1,⇒−1+1−1=−1 3 of theme−1−1−1=−3 0 of theme,1+1+1=3$
	$y e a h w i t h e p l e a s u r$ $s i n c e e v r e y a^{3} = \underline{+} (9)$ $w h a t i s p o s s i b l e v a l u e o f$ $x + y + z w i t h s z, y, x \in {- 1, 1}$ $i f o n e = - 1 \Rightarrow - 1 + 1 + 1 = 1$ $i f t w o = - 1, \Rightarrow - 1 + 1 - 1 = - 1$ $3 o f t h e m e - 1 - 1 - 1 = - 3$ $0 o f t h e m e, 1 + 1 + 1 = 3$
	Commented by mathocean1 last updated on 25/Oct/20

	$thank you very much sir .$
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