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Question Number 23050 by jazary last updated on 25/Oct/17

$$howcandemonstredthat$$$$\mathrm{\forall }a,b,c\in \mathbb{N}$$$$a^2+b^2=c^2\Rightarrow$$$$abc\equiv 0\left[60\right]$$

Answered by Rasheed.Sindhi last updated on 29/Oct/17

$$abc\equiv 0\left[60\right]\Rightarrow 60\mid abc$$$$\Rightarrow (4.3.5)\mid abc$$$$\Rightarrow 4\mid abc\wedge 3\mid abc\wedge 5\mid abc$$$${}^{\bullet }4\mid abc\Rightarrow 4\mid ab\vee 4\mid bc\vee 4\mid ac......\left(\mathrm{i}\right)$$$${}^{\bullet }3\mid abc\Rightarrow 3\mid a\vee 3\mid b\vee 3\mid c...........\left(\mathrm{ii}\right)$$$${}^{\bullet }5\mid abc\Rightarrow 5\mid a\vee 5\mid b\vee 5\mid c..........\left(\mathrm{iii}\right)$$$$\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{prove}:$$$$\mathrm{Case}-1:\left(\mathrm{i}\right):4\nmid ab\wedge 4\nmid bc\Rightarrow 4\mid ac$$$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{proved}\mathrm{like}\mathrm{Case}-2$$$$\mathrm{Case}-2:\left(\mathrm{ii}\right):3\nmid a\wedge 3\nmid b\Rightarrow 3\mid c$$$$3\nmid a\wedge 3\nmid b\mathrm{means}a\mathrm{and}b\mathrm{or}\mathrm{of}$$$$3\mathrm{k}+1\mathrm{or}3\mathrm{k}+2\mathrm{type}$$$$\Rightarrow a^2\mathrm{and}{b}^2\mathrm{are}\mathrm{of}3\mathrm{k}+1\mathrm{type}$$$$\mathrm{c}=\sqrt{a^2+b^2}=\sqrt{(3\mathrm{k}+1)+(3l+1)}$$$$=\sqrt{3\mathrm{m}+2}$$$$\mathrm{But}3\mathrm{m}+2\mathrm{is}\mathrm{not}\mathrm{perfect}\mathrm{square}$$$$\therefore \mathrm{If}\mathrm{c}\in \mathbb{N}\Rightarrow a^2+b^2\mathrm{is}\mathrm{perfect}\mathrm{square}$$$$\Rightarrow a^2\mathrm{and}{b}^2\mathrm{are}\mathrm{not}\mathrm{both}\mathrm{of}\mathrm{type}$$$$3\mathrm{k}+1.$$$$\therefore a\mathrm{or}b{\mathrm{aren}}^{\prime }\mathrm{t}\mathrm{of}\mathrm{type}3\mathrm{k}+1\mathrm{or}$$$$3\mathrm{k}+2.$$$$\therefore a\mathrm{or}b\mathrm{is}3\mathrm{k}-\mathrm{type}$$$$\therefore 3\mid a\mathrm{or}3\mid b$$$$\therefore 3\mid abc$$$$\mathrm{Case}-3:\left(\mathrm{iii}\right):5\nmid a\wedge 5\nmid b\Rightarrow 5\mid c$$$$\mathrm{This}\mathrm{case}\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{proved}\mathrm{like}\mathrm{Case}-2$$$$4\mid abc\mathrm{Case}-1$$$$3\mid abc\mathrm{Case}-2$$$$5\mid abc\mathrm{Case}-3$$$$\mathrm{Hence}60\mid abc$$$$\mathrm{Or}abc\equiv 0\left[60\right]$$

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