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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

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 .