show-that-if-a-2-b-2-can-be-divised-by-7-a-b-can-also-be-divised-by-7-

Question Number 118391 by mathocean1 last updated on 17/Oct/20

s h o w t h a t i f^{} a^{2} + b^{2} c a n b e d i v i s e d

b y 7, a + b c a n a l s o b e d i v i s e d b y 7 .

Answered by mindispower last updated on 17/Oct/20

a^{2} + b^{2} = 0 (7)

\Rightarrow a^{2} = - b^{2} (7)

\Rightarrow a^{2} = 6 b^{2} (7)

\Rightarrow b^{2} = (0, 1, 4, 2,) (7)

a^{2} = (0, 3, 5, 6)

a^{2} = 0 \Rightarrow a = 0

\Rightarrow b = 0 \Rightarrow a + b = 0 (7)

Answered by 1549442205PVT last updated on 17/Oct/20

We prove the stronger assert that :

a^{2} + b^{2} divisible by 7 if and only if a and

b are divisible by 7 simultaneously

Indeed,

\forall a, b \in Z we have a = 7 p + r_{1}, b = 7 q + r_{2}

with r_{i} \in {0, \pm 1, \pm 2, \pm 3} (i = 1, 2); p, q \in Z

Hence, a^{2} + b^{2} = {(7 p + r_{1})}^{2} + {(7 q + r_{2})}^{2}

= 49 (p^{2} + q^{2}) + 14 ({pr}_{1} + {qr}_{2}) + r_{1}^{2} + r_{2}^{2}

\Rightarrow (a^{2} + b^{2}) ⋮ 7 \Leftrightarrow (r_{1}^{2} + r_{2}^{2}) ⋮ 7 (*)

Since r_{i} \in {0, \pm 1, \pm 2, \pm 3} (i = 1, 2), we get

r_{i}^{2} \in {0, 1, 4, 9} \Rightarrow (r_{1}^{2} + r_{2}^{2}) \in {0, 1, 2, 16, 17,

, 32, 81, 82, 97, 162} \Rightarrow (r_{1}^{2} + r_{2}^{2}) ⋮ 7

\Leftrightarrow r_{1}^{2} + r_{2}^{2} = 0 \Leftrightarrow r_{1} = r_{2} = 0 \Leftrightarrow a = 7 p, b = 7 q

Therefore, a^{2} + b^{2} ⋮ 7 if and only if

a and b divisible by 7 simultaneously

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