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Question Number 118391 by mathocean1 last updated on 17/Oct/20
show that if ^ a^2 +b^2 can be divised by 7, a+b can also be divised by 7.
$${show}\:{that}\:{if}\:\:^{} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{can}\:{be}\:{divised} \\ $$$${by}\:\mathrm{7},\:{a}+{b}\:{can}\:{also}\:{be}\:{divised}\:{by}\:\mathrm{7}. \\ $$
Answered by mindispower last updated on 17/Oct/20
a^2 +b^2 =0(7) ⇒a^2 =−b^2 (7) ⇒a^2 =6b^2 (7) ⇒b^2 =(0,1,4,2,)(7) a^2 =(0,3,5,6) a^2 =0⇒a=0 ⇒b=0⇒a+b=0(7)
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0}\left(\mathrm{7}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} =−{b}^{\mathrm{2}} \left(\mathrm{7}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{6}{b}^{\mathrm{2}} \left(\mathrm{7}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\left(\mathrm{0},\mathrm{1},\mathrm{4},\mathrm{2},\right)\left(\mathrm{7}\right) \\ $$$${a}^{\mathrm{2}} =\left(\mathrm{0},\mathrm{3},\mathrm{5},\mathrm{6}\right) \\ $$$${a}^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{0}\Rightarrow{a}+{b}=\mathrm{0}\left(\mathrm{7}\right) \\ $$
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)
$$\mathrm{We}\:\mathrm{prove}\:\mathrm{the}\:\:\mathrm{stronger}\:\mathrm{assert}\:\mathrm{that}: \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:\mathrm{a}\:\mathrm{and} \\ $$$$\:\mathrm{b}\:\mathrm{are}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\:\mathrm{simultaneously} \\ $$$$\mathrm{Indeed}, \\ $$$$\forall\mathrm{a},\mathrm{b}\in\mathrm{Z}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}=\mathrm{7p}+\mathrm{r}_{\mathrm{1}} ,\mathrm{b}=\mathrm{7q}+\mathrm{r}_{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{r}_{\mathrm{i}} \in\left\{\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3}\right\}\left(\mathrm{i}=\mathrm{1},\mathrm{2}\right);\mathrm{p},\mathrm{q}\in\mathrm{Z} \\ $$$$\mathrm{Hence},\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\left(\mathrm{7p}+\mathrm{r}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{7q}+\mathrm{r}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$=\mathrm{49}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} \right)+\mathrm{14}\left(\mathrm{pr}_{\mathrm{1}} +\mathrm{qr}_{\mathrm{2}} \right)+\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\vdots\mathrm{7}\Leftrightarrow\left(\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} \right)\vdots\mathrm{7}\left(\ast\right) \\ $$$$\mathrm{Since}\:\:\mathrm{r}_{\mathrm{i}} \in\left\{\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3}\right\}\left(\mathrm{i}=\mathrm{1},\mathrm{2}\right),\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{r}_{\mathrm{i}} ^{\mathrm{2}} \in\left\{\mathrm{0},\mathrm{1},\mathrm{4},\mathrm{9}\right\}\Rightarrow\left(\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} \right)\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{16},\mathrm{17},\right. \\ $$$$\left.,\mathrm{32},\mathrm{81},\mathrm{82},\mathrm{97},\mathrm{162}\right\}\Rightarrow\left(\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} \right)\vdots\mathrm{7} \\ $$$$\Leftrightarrow\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{0}\Leftrightarrow\mathrm{r}_{\mathrm{1}} =\mathrm{r}_{\mathrm{2}} =\mathrm{0}\Leftrightarrow\mathrm{a}=\mathrm{7p},\mathrm{b}=\mathrm{7q} \\ $$$$\mathrm{Therefore},\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \vdots\mathrm{7}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\:\mathrm{simultaneously} \\ $$$$\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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