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

how can demonstred that      ∀a,b,c∈N   a^2 +b^2 =c^2   ⇒  abc≡0[60]

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

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

abc≡0[60]⇒60∣abc    ⇒(4.3.5)∣abc  ⇒4∣abc ∧ 3∣abc ∧ 5∣abc  ^• 4∣abc⇒4∣ab ∨ 4∣bc ∨ 4∣ ac......(i)  ^• 3∣abc⇒3∣a ∨ 3∣b ∨ 3∣c...........(ii)  ^• 5∣abc⇒5∣a ∨ 5∣b ∨ 5∣c..........(iii)  We have to prove:  Case-1:(i):4∤ab ∧ 4∤bc⇒4∣ac     This can be proved like Case-2  Case-2:(ii):3∤a ∧ 3∤b⇒3∣c  3∤a ∧ 3∤b means a and b or of  3k+1 or  3k+2 type  ⇒a^2  and b^2  are of 3k+1 type  c=(√(a^2 +b^2  ))  =(√((3k+1)+(3l+1)))     =(√(3m+2))  But 3m+2 is not perfect square  ∴ If  c∈N⇒a^2 +b^2  is perfect square  ⇒a^2 and b^2  are not both of type       3k+1.  ∴ a or b aren′t of type 3k+1 or  3k+2.  ∴ a or b is 3k-type  ∴ 3∣a or 3∣b  ∴3∣abc  Case-3:(iii):5∤a ∧ 5∤b⇒5∣c    This case can also be proved like Case-2  4∣abc   Case-1  3∣abc   Case-2  5∣abc    Case-3  Hence 60∣abc  Or abc≡0[60]

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

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