show that if ^ a^2 +b^2 can be divised by 7, a+b can also be divised by 7.


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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
	$We prove the stronger assert that: a^2 +b^2 divisible by 7 if and only if a and b are divisible by 7 simultaneously Indeed, ∀a,b∈Z we have a=7p+r_1 ,b=7q+r_2 with r_i ∈{0,±1,±2,±3}(i=1,2);p,q∈Z Hence,a^2 +b^2 =(7p+r_1 )^2 +(7q+r_2 )^2 =49(p^2 +q^2 )+14(pr_1 +qr_2 )+r_1 ^2 +r_2 ^2 ⇒(a^2 +b^2 )⋮7⇔(r_1 ^2 +r_2 ^2 )⋮7(∗) Since r_i ∈{0,±1,±2,±3}(i=1,2),we get r_i ^2 ∈{0,1,4,9}⇒(r_1 ^2 +r_2 ^2 )∈{0,1,2,16,17, ,32,81,82,97,162}⇒(r_1 ^2 +r_2 ^2 )⋮7 ⇔r_1 ^2 +r_2 ^2 =0⇔r_1 =r_2 =0⇔a=7p,b=7q Therefore,a^2 +b^2 ⋮7 if and only if a and b divisible by 7 simultaneously (q.e.d)$
	$We prove the stronger assert that :$ $a^{2} + b^{2} divisible by 7 if and only if a and$ $b are divisible by 7 simultaneously$ $Indeed,$ $\forall a, b \in Z we have a = 7 p + r_{1}, b = 7 q + r_{2}$ $with r_{i} \in {0, \pm 1, \pm 2, \pm 3} (i = 1, 2); p, q \in Z$ $Hence, a^{2} + b^{2} = {(7 p + r_{1})}^{2} + {(7 q + r_{2})}^{2}$ $= 49 (p^{2} + q^{2}) + 14 ({pr}_{1} + {qr}_{2}) + r_{1}^{2} + r_{2}^{2}$ $\Rightarrow (a^{2} + b^{2}) ⋮ 7 \Leftrightarrow (r_{1}^{2} + r_{2}^{2}) ⋮ 7 (*)$ $Since r_{i} \in {0, \pm 1, \pm 2, \pm 3} (i = 1, 2), we get$ $r_{i}^{2} \in {0, 1, 4, 9} \Rightarrow (r_{1}^{2} + r_{2}^{2}) \in {0, 1, 2, 16, 17,$ $, 32, 81, 82, 97, 162} \Rightarrow (r_{1}^{2} + r_{2}^{2}) ⋮ 7$ $\Leftrightarrow r_{1}^{2} + r_{2}^{2} = 0 \Leftrightarrow r_{1} = r_{2} = 0 \Leftrightarrow a = 7 p, b = 7 q$ $Therefore, a^{2} + b^{2} ⋮ 7 if and only if$ $a and b divisible by 7 simultaneously$ $(q . e . d)$
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